Now, the nice thing about working with infinities is that you don't have to be very precise at all, and you'll still get exactly the right answer at the end!

Instead of taking x^N, let's use the 'rising powers', okay that didn't translate so probably is not a thing in Golarion, or at least the Share Language donor didn't know it.

Instead of taking x * x * x * ... * x, with N copies of x, let's consider the expression x * (x + 1) * (x + 2) ... until you've got N of those.

Let's denote that x`N`.

Obviously if x is any finite fraction of the way towards infinity, x`N` = x^N exactly.  Like, when you're working with trillions, a trillion squared is nearly the same number as a trillion times a trillion plus one, proportionally speaking; that's approximately true, so with infinities it's exactly true.

For example with x=10, N=3:

10`3` = 10*11*12

Now observe that:

11`3` - 10`3`
= 11*12*13 - 10*11*12
= (13 - 10) * (11 * 12)
= 3 * 11`2`

By obvious generalization:

(x + 1)`N` - (x)`N`
= (x + 1)(x + 2)...(x + N) - (x)(x+1)(x+2)...(x + N - 1)
= (x + N - x) * (x+1)(x+2)...(x + N - 1)
= N * (x+1)`N - 1`

Rearrange:

x`N` + N*(x+1)`N-1` = (x+1)`N`

Eliminate all subtraction whenever possible, in case something ends up 0 or negative that didn't start out that way:

x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

Sanity check with x=10, N=2:

10*11*12 + 3*(11*12) = 11*12*13:   PASSED

Induction, starting from N * 0`N` = 0`N + 1` = 0:

(N+1) * 0`N`  +  (N+1) * 1`N`  +  (N+1) * 2`N` + ... (N+1) * x`N`   =   x`N + 1`

Sanity check with x=3, N = 2:

3(0)(1) + 3(1)(2) + 3(2)(3) + 3(3)(4) = (3)(4)(5)
0 + 6 + 18 + 36 = 60 

Divide both sides by N+1:

0`N` + 1`N` + 2`N` + 3`N` + ... + x`N` = x`N+1` / (N + 1)

Approximations are exactly correct at infinity:

(INF)`N` = (INF)^N

...and this is also exactly true once you get any substantial fraction of the way towards infinity.  Like, even (1,000,000)`3` starts to look pretty close to (1,000,000)^3.

As for all the finite numbers at the start of the sequence, they don't contribute anything and can be ignored.

Conclusion:

0^N + 1^N + 2^N + 3^N + ... + INF^N = INF^(N+1) / (N+1).

Now, the nice thing about working with infinities is that you don't have to be very precise at all, and you'll still get exactly the right answer at the end!

Instead of taking x^N, let's use the 'rising powers', okay that didn't translate so probably is not a thing in Golarion, or at least the Share Language donor didn't know it.

Instead of taking x * x * x * ... * x, with N copies of x, let's consider the expression x * (x + 1) * (x + 2) ... until you've got N of those.

Let's denote that x`N`.

Obviously if x is any finite fraction of the way towards infinity, x`N` = x^N exactly.  Like, when you're working with trillions, a trillion squared is nearly the same number as a trillion times a trillion plus one, proportionally speaking; that's approximately true, so with infinities it's exactly true.

Example with x=10, N=3:

10`3` = 10*11*12

Now observe that:

11`3` - 10`3`
= 11*12*13 - 10*11*12
= (13 - 10) * (11 * 12)
= 3 * 11`2`

By obvious generalization:

(x + 1)`N` - (x)`N`
= (x + 1)(x + 2)...(x + N) - (x)(x+1)(x+2)...(x + N - 1)
= (x + N - x) * (x+1)(x+2)...(x + N - 1)
= N * (x+1)`N - 1`

Rearrange:

x`N` + N*(x+1)`N-1` = (x+1)`N`

Eliminate all subtraction whenever possible, in case something ends up 0 or negative that didn't start out that way:

x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

Sanity check with x=10, N=2:

10*11*12 + 3*(11*12) = 11*12*13:   PASSED

Induction, starting from N * 0`N` = 0`N + 1` = 0:

(N+1) * 0`N`  +  (N+1) * 1`N`  +  (N+1) * 2`N` + ... (N+1) * x`N`   =   x`N + 1`

Sanity check with x=3, N = 2:

3(0)(1) + 3(1)(2) + 3(2)(3) + 3(3)(4) = (3)(4)(5)
0 + 6 + 18 + 36 = 60
PASSED

Divide both sides by N+1:

0`N` + 1`N` + 2`N` + 3`N` + ... + x`N` = x`N+1` / (N + 1)

Approximations are exactly correct at infinity:

(INF)`N` = (INF)^N

...and this is also exactly true once you get any substantial fraction of the way towards infinity.  Like, even (1,000,000)`3` starts to look pretty close to (1,000,000)^3.

As for all the finite numbers at the start of the sequence, they don't contribute anything and can be ignored.

Conclusion:

0^N + 1^N + 2^N + 3^N + ... + INF^N = INF^(N+1) / (N+1).

Now, the nice thing about working with infinities is that you don't have to be very precise at all, and you'll still get exactly the right answer at the end!

Instead of taking x^N, let's use the 'rising powers', okay that didn't translate so probably is not a thing in Golarion, or at least the Share Language donor didn't know it.

Instead of taking x * x * x * ... * x, with N copies of x, let's consider the expression x * (x + 1) * (x + 2) ... until you've got N of those.

Let's denote that x`N`.

Obviously if x is any finite fraction of the way towards infinity, x`N` = x^N exactly.  Like, when you're working with trillions, a trillion squared is nearly the same number as a trillion times a trillion plus one, proportionally speaking; that's approximately true, so with infinities it's exactly true.

Example with x=10, N=3:

10`3` = 10*11*12

Now observe that:

11`3` - 10`3`
= 11*12*13 - 10*11*12
= (13 - 10) * (11 * 12)
= 3 * 11`2`

By obvious generalization:

(x + 1)`N` - (x)`N`
= (x + 1)(x + 2)...(x + N) - (x)(x+1)(x+2)...(x + N - 1)
= (x + N - x) * (x+1)(x+2)...(x + N - 1)
= N * (x+1)`N - 1`

Rearrange:

x`N` + N*(x+1)`N-1` = (x+1)`N`

Eliminate all subtraction whenever possible, in case something ends up 0 or negative that didn't start out that way:

x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

Sanity check with x=10, N=2:

10*11*12 + 3*(11*12) = 11*12*13:   PASSED

Induction, starting from N * 0`N` = 0`N + 1` = 0:

(N+1) * 0`N`  +  (N+1) * 1`N`  +  (N+1) * 2`N` + ... (N+1) * x`N`   =   x`N + 1`

Sanity check with x=3, N = 2:

3(0)(1) + 3(1)(2) + 3(2)(3) + 3(3)(4) = (3)(4)(5)
0 + 6 + 18 + 36 = 60
PASSED

Divide both sides by N+1:

0`N` + 1`N` + 2`N` + 3`N` + ... + x`N` = x`N+1` / (N + 1)

Approximations are exactly correct at infinity:

(INF)`N` = (INF)^N

...and this is also exactly true once you get any substantial fraction of the way towards infinity.  Like, even (1,000,000)`3` starts to look pretty close to (1,000,000)^3.

As for all the finite numbers at the start of the sequence, they're collectively 0% of anything and can be ignored.

Conclusion:

0^N + 1^N + 2^N + 3^N + ... + INF^N = INF^(N+1) / (N+1).

Now, the nice thing about working with infinities is that you don't have to be very precise at all, and you'll still get exactly the right answer at the end!

Instead of taking x^N, let's use the 'rising powers', okay that didn't translate so probably is not a thing in Golarion, or at least the Share Language donor didn't know it.

Instead of taking x * x * x * ... * x, with N copies of x, let's consider the expression x * (x + 1) * (x + 2) ... until you've got N of those.

Let's denote that x`N`.

Obviously if x is any finite fraction of the way towards infinity, x`N` = x^N exactly.  Like, when you're working with trillions, a trillion squared is nearly the same number as a trillion times a trillion plus one, proportionally speaking; that's approximately true, so with infinities it's exactly true.

Example with x=10, N=3:

10`3` = 10*11*12

Now observe that:

11`3` - 10`3`
= 11*12*13 - 10*11*12
= (13 - 10) * (11 * 12)
= 3 * 11`2`

By obvious generalization:

(x + 1)`N` - (x)`N`
= (x + 1)(x + 2)...(x + N) - (x)(x+1)(x+2)...(x + N - 1)
= (x + N - x) * (x+1)(x+2)...(x + N - 1)
= N * (x+1)`N - 1`

Rearrange:

x`N` + N*(x+1)`N-1` = (x+1)`N`

Eliminate all subtraction whenever possible, in case something ends up 0 or negative that didn't start out that way:

x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

Sanity check with x=10, N=2:

10`2+1` + (2+1)*(10+1)`2` = (10+1)`2+1`
=> 10*11*12 + 3*(11*12) = 11*12*13
PASSED

Induction, starting from N * 0`N` = 0`N + 1` = 0:

(N+1) * 0`N`  +  (N+1) * 1`N`  +  (N+1) * 2`N` + ... (N+1) * x`N`   =   x`N + 1`

Sanity check with x=3, N = 2:

3(0)(1) + 3(1)(2) + 3(2)(3) + 3(3)(4) = (3)(4)(5)
0 + 6 + 18 + 36 = 60
PASSED

Divide both sides by N+1:

0`N` + 1`N` + 2`N` + 3`N` + ... + x`N` = x`N+1` / (N + 1)

Approximations are exactly correct at infinity:

(INF)`N` = (INF)^N

...and this is also exactly true once you get any substantial fraction of the way towards infinity.  Like, even (1,000,000)`3` starts to look pretty close to (1,000,000)^3.

As for all the finite numbers at the start of the sequence, they collectively comprise 0% of anything and can be ignored.

Conclusion:

0^N + 1^N + 2^N + 3^N + ... + INF^N = INF^(N+1) / (N+1).

Now, the nice thing about working with infinities is that you don't have to be very precise at all, and you'll still get exactly the right answer at the end!

Instead of taking x^N, let's use the 'rising powers', okay that didn't translate so probably is not a thing in Golarion, or at least the Share Language donor didn't know it.

Instead of taking x * x * x * ... * x, with N copies of x, let's consider the expression x * (x + 1) * (x + 2) ... until you've got N of those.

Let's denote that x`N`.

Obviously if x is any finite fraction of the way towards infinity, x`N` = x^N exactly.  Like, when you're working with trillions, a trillion squared is nearly the same number as a trillion times a trillion plus one, proportionally speaking; that's approximately true, so with infinities it's exactly true.

Example with x=10, N=3:

10`3` = 10*11*12

Now observe that:

11`3` - 10`3`
= 11*12*13 - 10*11*12
= (13 - 10) * (11 * 12)
= 3 * 11`2`

By obvious generalization:

(x + 1)`N` - (x)`N`
= (x + 1)(x + 2)...(x + N) - (x)(x+1)(x+2)...(x + N - 1)
= (x + N - x) * (x+1)(x+2)...(x + N - 1)
= N * (x+1)`N - 1`

Rearrange:

x`N` + N*(x+1)`N-1` = (x+1)`N`

Eliminate all subtraction whenever possible, in case something ends up 0 or negative that didn't start out that way:

x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

Sanity check with x=10, N=2:

10`2+1` + (2+1)*(10+1)`2` = (10+1)`2+1`
=> 10*11*12 + 3*(11*12) = 11*12*13
PASSED

Induction, starting from N * 0`N` = 0`N + 1` = 0:

(N+1) * 0`N`  +  (N+1) * 1`N`  +  (N+1) * 2`N` + ... (N+1) * x`N`   =   x`N + 1`

Sanity check with x=3, N = 2:

3(0)(1) + 3(1)(2) + 3(2)(3) + 3(3)(4) = (3)(4)(5)
0 + 6 + 18 + 36 = 60
PASSED

Divide both sides by N+1:

0`N` + 1`N` + 2`N` + 3`N` + ... + x`N` = x`N+1` / (N + 1)

Approximations are exactly correct at infinity:

(INF)`N` = (INF)^N

...and this is also exactly true once you get any substantial fraction of the way towards infinity.  Like, even (1,000,000)`3` starts to look pretty close to (1,000,000)^3.

As for all the finite numbers at the start of the sequence, they collectively comprise 0% of anything and can be ignored.

Conclusion:

0^N + 1^N + 2^N + 3^N + ... + INF^N = INF^(N+1) / (N+1).

So if you stack up an infinite number of squares with sides 0, 1, 2, 3... you will get exactly a third of an infinitely large cube, and more generally, if you stack up an infinite number of N-hypercubes with sides 0, 1, 2, 3... you'll get exactly 1 / (N+1) of an infinite (N+1)-hypercube.

Now, the nice thing about working with infinities is that you don't have to be very precise at all, and you'll still get exactly the right answer at the end!

Instead of taking x^N, let's use the 'rising powers', okay that didn't translate so probably is not a thing in Golarion, or at least the Share Language donor didn't know it.

Instead of taking x * x * x * ... * x, with N copies of x, let's consider the expression x * (x + 1) * (x + 2) ... until you've got N of those.

Let's denote that x`N`.

Obviously if x is any finite fraction of the way towards infinity, x`N` = x^N exactly.  Like, when you're working with trillions, a trillion squared is nearly the same number as a trillion times a trillion plus one, proportionally speaking; that's approximately true, so with infinities it's exactly true.

Example with x=10, N=3:

10`3` = 10*11*12

Now observe that:

11`3` - 10`3`
= 11*12*13 - 10*11*12
= (13 - 10) * (11 * 12)
= 3 * 11`2`

By obvious generalization:

(x + 1)`N` - (x)`N`
= (x + 1)(x + 2)...(x + N) - (x)(x+1)...(x + N - 1)
= (x + N - x) * (x+1)(x+2)...(x + N - 1)
= N * (x+1)`N - 1`

Rearrange:

x`N` + N*(x+1)`N-1` = (x+1)`N`

Eliminate all subtraction whenever possible, in case something ends up 0 or negative that didn't start out that way:

x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

Sanity check with x=10, N=2:

10`2+1` + (2+1)*(10+1)`2` = (10+1)`2+1`
=> 10*11*12 + 3*(11*12) = 11*12*13
PASSED

Induction, starting from N * 0`N` = 0`N + 1` = 0:

(N+1) * 0`N`  +  (N+1) * 1`N`  +  (N+1) * 2`N` + ... (N+1) * x`N`   =   x`N + 1`

Sanity check with x=3, N = 2:

3(0)(1) + 3(1)(2) + 3(2)(3) + 3(3)(4) = (3)(4)(5)
0 + 6 + 18 + 36 = 60
PASSED

Divide both sides by N+1:

0`N` + 1`N` + 2`N` + 3`N` + ... + x`N` = x`N+1` / (N + 1)

Approximations are exactly correct at infinity:

(INF)`N` = (INF)^N

...and this is also exactly true once you get any substantial fraction of the way towards infinity.  Like, even (1,000,000)`3` starts to look pretty close to (1,000,000)^3.

As for all the finite numbers at the start of the sequence, they collectively comprise 0% of anything and can be ignored.

Conclusion:

0^N + 1^N + 2^N + 3^N + ... + INF^N = INF^(N+1) / (N+1).

So if you stack up an infinite number of squares with sides 0, 1, 2, 3... you will get exactly a third of an infinitely large cube, and more generally, if you stack up an infinite number of N-hypercubes with sides 0, 1, 2, 3... you'll get exactly 1 / (N+1) of an infinite (N+1)-hypercube.

Now, the nice thing about working with infinities is that you don't have to be very precise at all, and you'll still get exactly the right answer at the end!

Instead of taking x^N, let's use the 'rising powers', okay that didn't translate so probably is not a thing in Golarion, or at least the Share Language donor didn't know it.

Instead of taking x * x * x * ... * x, with N copies of x, let's consider the expression x * (x + 1) * (x + 2) ... until you've got N of those.

Let's denote that x`N`.

Obviously if x is any finite fraction of the way towards infinity, x`N` = x^N exactly.  Like, when you're working with trillions, a trillion squared is nearly the same number as a trillion times a trillion plus one, proportionally speaking; that's approximately true, so with infinities it's exactly true.

Example with x=10, N=3:

10`3` = 10*11*12

Now observe that:

11`3` - 10`3`
= 11*12*13 - 10*11*12
= (13 - 10) * (11 * 12)
= 3 * 11`2`

By obvious generalization:

   (x + 1)`N`  -  (x)`N`
= (x + 1)(x + 2)...(x + N)  -  (x)(x+1)...(x + N - 1)
= (x + N - x) * (x+1)(x+2)...(x + N - 1)
= N * (x+1)`N - 1`

Rearrange:

x`N` + N*(x+1)`N-1` = (x+1)`N`

Eliminate all subtraction whenever possible, in case something ends up 0 or negative that didn't start out that way:

x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

Sanity check with x=10, N=2:

10`2+1` + (2+1)*(10+1)`2` = (10+1)`2+1`
=> 10*11*12 + 3*(11*12) = 11*12*13
PASSED

Induction, starting from N * 0`N` = 0`N + 1` = 0:

(N+1) * 0`N`  +  (N+1) * 1`N`  +  (N+1) * 2`N` + ... (N+1) * x`N`   =   x`N + 1`

Sanity check with x=3, N = 2:

3(0)(1) + 3(1)(2) + 3(2)(3) + 3(3)(4) = (3)(4)(5)
0 + 6 + 18 + 36 = 60
PASSED

Divide both sides by N+1:

0`N` + 1`N` + 2`N` + 3`N` + ... + x`N` = x`N+1` / (N + 1)

Approximations are exactly correct at infinity:

(INF)`N` = (INF)^N

...and this is also exactly true once you get any substantial fraction of the way towards infinity.  Like, even (1,000,000)`3` starts to look pretty close to (1,000,000)^3.

As for all the finite numbers at the start of the sequence, they collectively comprise 0% of anything and can be ignored.

Conclusion:

0^N + 1^N + 2^N + 3^N + ... + INF^N = INF^(N+1) / (N+1).

So if you stack up an infinite number of squares with sides 0, 1, 2, 3... you will get exactly a third of an infinitely large cube, and more generally, if you stack up an infinite number of N-hypercubes with sides 0, 1, 2, 3... you'll get exactly 1 / (N+1) of an infinite (N+1)-hypercube.

Now, the nice thing about working with infinities is that you don't have to be very precise at all, and you'll still get exactly the right answer at the end!

Instead of taking x^N, let's use the 'rising powers', okay that didn't translate so probably is not a thing in Golarion, or at least the Share Language donor didn't know it.

Instead of taking x * x * x * ... * x, with N copies of x, let's consider the expression x * (x + 1) * (x + 2) ... until you've got N of those.

Let's denote that x`N`.

Obviously if x is any finite fraction of the way towards infinity, x`N` = x^N exactly.  Like, when you're working with trillions, a trillion squared is nearly the same number as a trillion times a trillion plus one, proportionally speaking; that's approximately true, so with infinities it's exactly true.

Example with x=10, N=3:

10`3` = 10*11*12

Now observe that:

11`3` - 10`3`
= 11*12*13 - 10*11*12
= (13 - 10) * (11 * 12)
= 3 * 11`2`

By pointwise generalization from 10 to x and 3 to N:

   (x + 1)`N`  -  (x)`N`
= (x + 1)(x + 2)...(x + N)  -  (x)(x+1)...(x + N - 1)
= ((x + N) - x) * (x+1)(x+2)...(x + N - 1)
= N * (x+1)`N - 1`

Rearrange:

x`N` + N*(x+1)`N-1` = (x+1)`N`

Eliminate all subtraction whenever possible, in case something ends up 0 or negative that didn't start out that way:

x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

Sanity check with x=10, N=2:

10`2+1` + (2+1)*(10+1)`2` = (10+1)`2+1`
=> 10*11*12 + 3*(11*12) = 11*12*13
PASSED

Induction, starting from N * 0`N` = 0`N + 1` = 0:

(N+1) * 0`N`  +  (N+1) * 1`N`  +  (N+1) * 2`N` + ... (N+1) * x`N`   =   x`N + 1`

Sanity check with x=3, N = 2:

3(0)(1) + 3(1)(2) + 3(2)(3) + 3(3)(4) = (3)(4)(5)
0 + 6 + 18 + 36 = 60
PASSED

Divide both sides by N+1:

0`N` + 1`N` + 2`N` + 3`N` + ... + x`N` = x`N+1` / (N + 1)

Approximations are exactly correct at infinity:

(INF)`N` = (INF)^N

...and this is also exactly true once you get any substantial fraction of the way towards infinity.  Like, even (1,000,000)`3` starts to look pretty close to (1,000,000)^3.

As for all the finite numbers at the start of the sequence, they collectively comprise 0% of anything and can be ignored.

Conclusion:

0^N + 1^N + 2^N + 3^N + ... + INF^N = INF^(N+1) / (N+1).

So if you stack up an infinite number of squares with sides 0, 1, 2, 3... you will get exactly a third of an infinitely large cube, and more generally, if you stack up an infinite number of N-hypercubes with sides 0, 1, 2, 3... you'll get exactly 1 / (N+1) of an infinite (N+1)-hypercube.

Now, the nice thing about working with infinities is that you don't have to be very precise at all, and you'll still get exactly the right answer at the end!

Instead of taking x^N, let's use the 'rising powers', okay that didn't translate so probably is not a thing in Golarion, or at least the Share Language donor didn't know it.

Instead of taking x * x * x * ... * x, with N copies of x, let's consider the expression x * (x + 1) * (x + 2) ... until you've got N of those.

Let's denote that x`N`.

Obviously if x is any finite fraction of the way towards infinity, x`N` = x^N exactly.  Like, when you're working with trillions, a trillion squared is nearly the same number as a trillion times a trillion plus one, proportionally speaking; that's approximately true, so with infinities it's exactly true.

Example with x=10, N=3:

10`3` = 10*11*12

Now observe that:

11`3` - 10`3`
= 11*12*13 - 10*11*12
= (13 - 10) * (11 * 12)
= 3 * 11`2`

By pointwise-exact generalization from 10 to x and 3 to N:

   (x + 1)`N`  -  (x)`N`
= (x + 1)(x + 2)...(x + N)  -  (x)(x+1)...(x + N - 1)
= ((x + N) - x) * (x+1)(x+2)...(x + N - 1)
= N * (x+1)`N - 1`

Rearrange:

x`N` + N*(x+1)`N-1` = (x+1)`N`

Eliminate all subtraction whenever possible, in case something ends up 0 or negative that didn't start out that way:

x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

Sanity check with x=10, N=2:

10`2+1` + (2+1)*(10+1)`2` = (10+1)`2+1`
=> 10*11*12 + 3*(11*12) = 11*12*13
PASSED

Induction, starting from N * 0`N` = 0`N + 1` = 0:

(N+1) * 0`N`  +  (N+1) * 1`N`  +  (N+1) * 2`N` + ... (N+1) * x`N`   =   x`N + 1`

Sanity check with x=3, N = 2:

3(0)(1) + 3(1)(2) + 3(2)(3) + 3(3)(4) = (3)(4)(5)
0 + 6 + 18 + 36 = 60
PASSED

Divide both sides by N+1:

0`N` + 1`N` + 2`N` + 3`N` + ... + x`N` = x`N+1` / (N + 1)

Approximations are exactly correct at infinity:

(INF)`N` = (INF)^N

...and this is also exactly true once you get any substantial fraction of the way towards infinity.  Like, even (1,000,000)`3` starts to look pretty close to (1,000,000)^3.

As for all the finite numbers at the start of the sequence, they collectively comprise 0% of anything and can be ignored.

Conclusion:

0^N + 1^N + 2^N + 3^N + ... + INF^N = INF^(N+1) / (N+1).

So if you stack up an infinite number of squares with sides 0, 1, 2, 3... you will get exactly a third of an infinitely large cube, and more generally, if you stack up an infinite number of N-hypercubes with sides 0, 1, 2, 3... you'll get exactly 1 / (N+1) of an infinite (N+1)-hypercube.

Now, the nice thing about working with infinities is that you don't have to be very precise at all, and you'll still get exactly the right answer at the end!

Instead of taking x^N, let's use the 'rising powers', okay that didn't translate so probably is not a thing in Golarion, or at least the Share Language donor didn't know it.

Instead of taking x * x * x * ... * x, with N copies of x, let's consider the expression x * (x + 1) * (x + 2) ... until you've got N of those.

Let's denote that x`N`.

Obviously if x is any finite fraction of the way towards infinity, x`N` = x^N exactly.  Like, when you're working with trillions, a trillion squared is nearly the same number as a trillion times a trillion plus one, proportionally speaking; that's approximately true, so with infinities it's exactly true.

Example with x=10, N=3:

10`3` = 10*11*12

Now observe that:

11`3` - 10`3`
= 11*12*13 - 10*11*12
= (13 - 10) * (11 * 12)
= 3 * 11`2`

By (pointwise exact term-by-term) generalization from 10 to x and 3 to N:

   (x + 1)`N`  -  (x)`N`
= (x + 1)(x + 2)...(x + N)  -  (x)(x+1)...(x + N - 1)
= ((x + N) - x) * (x+1)...(x + N - 1)
= N * (x+1)`N - 1`

Rearrange:

x`N` + N*(x+1)`N-1` = (x+1)`N`

Eliminate all subtraction whenever possible, in case something ends up 0 or negative that didn't start out that way:

x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

Sanity check with x=10, N=2:

10`2+1` + (2+1)*(10+1)`2` = (10+1)`2+1`
=> 10*11*12 + 3*(11*12) = 11*12*13
PASSED

Induction, starting from N * 0`N` = 0`N + 1` = 0:

(N+1) * 0`N`  +  (N+1) * 1`N`  +  (N+1) * 2`N` + ... (N+1) * x`N`   =   x`N + 1`

Sanity check with x=3, N = 2:

3(0)(1) + 3(1)(2) + 3(2)(3) + 3(3)(4) = (3)(4)(5)
0 + 6 + 18 + 36 = 60
PASSED

Divide both sides by N+1:

0`N` + 1`N` + 2`N` + 3`N` + ... + x`N` = x`N+1` / (N + 1)

Approximations are exactly correct at infinity:

(INF)`N` = (INF)^N

...and this is also exactly true once you get any substantial fraction of the way towards infinity.  Like, even (1,000,000)`3` starts to look pretty close to (1,000,000)^3.

As for all the finite numbers at the start of the sequence, they collectively comprise 0% of anything and can be ignored.

Conclusion:

0^N + 1^N + 2^N + 3^N + ... + INF^N = INF^(N+1) / (N+1).

So if you stack up an infinite number of squares with sides 0, 1, 2, 3... you will get exactly a third of an infinitely large cube, and more generally, if you stack up an infinite number of N-hypercubes with sides 0, 1, 2, 3... you'll get exactly 1 / (N+1) of an infinite (N+1)-hypercube.

Now, the nice thing about working with infinities is that you don't have to be very precise at all, and you'll still get exactly the right answer at the end!

Instead of taking x^N, let's use the 'rising powers', okay that didn't translate so probably is not a thing in Golarion, or at least the Share Language donor didn't know it.

Instead of taking x * x * x * ... * x, with N copies of x, let's consider the expression x * (x + 1) * (x + 2) ... until you've got N of those.

Let's denote that x`N`.

Obviously if x is any finite fraction of the way towards infinity, x`N` = x^N exactly.  Like, when you're working with trillions, a trillion squared is nearly the same number as a trillion times a trillion plus one, proportionally speaking; that's approximately true, so with infinities it's exactly true.

Example with x=10, N=3:

10`3`
= 10*11*12
= (10)(10+1)...(10+(N-1))

Now observe that:

11`3` - 10`3`
= 11*12*13 - 10*11*12
= (13 - 10) * (11 * 12)
= 3 * 11`2`

By (pointwise exact term-by-term) generalization from 10 to x and 3 to N:

   (x + 1)`N`  -  (x)`N`
= (x + 1)(x + 2)...(x + N)  -  (x)(x+1)...(x + N - 1)
= ((x + N) - x) * (x+1)...(x + N - 1)
= N * (x+1)`N - 1`

Rearrange:

x`N` + N*(x+1)`N-1` = (x+1)`N`

Eliminate all subtraction whenever possible, in case something ends up 0 or negative that didn't start out that way:

x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

Sanity check with x=10, N=2:

10`2+1` + (2+1)*(10+1)`2` = (10+1)`2+1`
=> 10*11*12 + 3*(11*12) = 11*12*13
PASSED

Induction, starting from N * 0`N` = 0`N + 1` = 0:

(N+1) * 0`N`  +  (N+1) * 1`N`  +  (N+1) * 2`N` + ... (N+1) * x`N`   =   x`N + 1`

Sanity check with x=3, N = 2:

3(0)(1) + 3(1)(2) + 3(2)(3) + 3(3)(4) = (3)(4)(5)
0 + 6 + 18 + 36 = 60
PASSED

Divide both sides by N+1:

0`N` + 1`N` + 2`N` + 3`N` + ... + x`N` = x`N+1` / (N + 1)

Approximations are exactly correct at infinity:

(INF)`N` = (INF)^N

...and this is also exactly true once you get any substantial fraction of the way towards infinity.  Like, even (1,000,000)`3` starts to look pretty close to (1,000,000)^3.

As for all the finite numbers at the start of the sequence, they collectively comprise 0% of anything and can be ignored.

Conclusion:

0^N + 1^N + 2^N + 3^N + ... + INF^N = INF^(N+1) / (N+1).

So if you stack up an infinite number of squares with sides 0, 1, 2, 3... you will get exactly a third of an infinitely large cube, and more generally, if you stack up an infinite number of N-hypercubes with sides 0, 1, 2, 3... you'll get exactly 1 / (N+1) of an infinite (N+1)-hypercube.

Now, the nice thing about working with infinities is that you don't have to be very precise at all, and you'll still get exactly the right answer at the end!

Instead of taking x^N, let's use the 'rising powers', okay that didn't translate so probably is not a thing in Golarion, or at least the Share Language donor didn't know it.

Instead of taking x * x * x * ... * x, with N copies of x, let's consider the expression x * (x + 1) * (x + 2) ... until you've got N of those.

Let's denote that x`N`.

Obviously if x is any finite fraction of the way towards infinity, x`N` = x^N exactly.  Like, when you're working with trillions, a trillion squared is nearly the same number as a trillion times a trillion plus one, proportionally speaking; that's approximately true, so with infinities it's exactly true.

Example with x=10, N=3:

10`3`
= 10*11*12
= (10)(10+1)...(10+(N-1))    (N terms total)

Now observe that:

11`3` - 10`3`
= 11*12*13 - 10*11*12
= (13 - 10) * (11 * 12)
= 3 * 11`2`

By (pointwise exact term-by-term) generalization from 10 to x and 3 to N:

   (x + 1)`N`  -  (x)`N`
= (x + 1)(x + 2)...(x + N)  -  (x)(x+1)...(x + N - 1)
= ((x + N) - x) * (x+1)...(x + N - 1)
= N * (x+1)`N - 1`

Rearrange:

x`N` + N*(x+1)`N-1` = (x+1)`N`

Eliminate all subtraction whenever possible, in case something ends up 0 or negative that didn't start out that way:

x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

Sanity check with x=10, N=2:

10`2+1` + (2+1)*(10+1)`2` = (10+1)`2+1`
=> 10*11*12 + 3*(11*12) = 11*12*13
PASSED

Induction, starting from N * 0`N` = 0`N + 1` = 0:

(N+1) * 0`N`  +  (N+1) * 1`N`  +  (N+1) * 2`N` + ... (N+1) * x`N`   =   x`N + 1`

Sanity check with x=3, N = 2:

3(0)(1) + 3(1)(2) + 3(2)(3) + 3(3)(4) = (3)(4)(5)
0 + 6 + 18 + 36 = 60
PASSED

Divide both sides by N+1:

0`N` + 1`N` + 2`N` + 3`N` + ... + x`N` = x`N+1` / (N + 1)

Approximations are exactly correct at infinity:

(INF)`N` = (INF)^N

...and this is also exactly true once you get any substantial fraction of the way towards infinity.  Like, even (1,000,000)`3` starts to look pretty close to (1,000,000)^3.

As for all the finite numbers at the start of the sequence, they collectively comprise 0% of anything and can be ignored.

Conclusion:

0^N + 1^N + 2^N + 3^N + ... + INF^N = INF^(N+1) / (N+1).

So if you stack up an infinite number of squares with sides 0, 1, 2, 3... you will get exactly a third of an infinitely large cube, and more generally, if you stack up an infinite number of N-hypercubes with sides 0, 1, 2, 3... you'll get exactly 1 / (N+1) of an infinite (N+1)-hypercube.

Now, the nice thing about working with infinities is that you don't have to be very precise at all, and you'll still get exactly the right answer at the end!

Instead of taking x^N, let's use the 'rising powers', okay that didn't translate so probably is not a thing in Golarion, or at least the Share Language donor didn't know it.

Instead of taking x * x * x * ... * x, with N copies of x, let's consider the expression x * (x + 1) * (x + 2) ... until you've got N of those.

Let's denote that x`N`.

Obviously if x is any finite fraction of the way towards infinity, x`N` = x^N exactly.  Like, when you're working with trillions, a trillion squared is nearly the same number as a trillion times a trillion plus one, proportionally speaking; that's approximately true, so with infinities it's exactly true.

Example with x=10, N=3:

10`3`
= 10*11*12
= (10)(10+1)...(10+(3-1))    (3 terms total)
= (x)(x+1)...(x+(N-1))          (x=10, N=3, N terms total)

Now observe that:

11`3` - 10`3`
= 11*12*13 - 10*11*12
= (13 - 10) * (11 * 12)
= 3 * 11`2`

By (pointwise exact term-by-term) generalization from 10 to x and 3 to N:

   (x + 1)`N`  -  (x)`N`
= (x + 1)(x + 2)...(x + N)  -  (x)(x+1)...(x + N - 1)
= ((x + N) - x) * (x+1)...(x + N - 1)
= N * (x+1)`N - 1`

Rearrange:

x`N` + N*(x+1)`N-1` = (x+1)`N`

Eliminate all subtraction whenever possible, in case something ends up 0 or negative that didn't start out that way:

x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

Sanity check with x=10, N=2:

10`2+1` + (2+1)*(10+1)`2` = (10+1)`2+1`
=> 10*11*12 + 3*(11*12) = 11*12*13
PASSED

Induction, starting from N * 0`N` = 0`N + 1` = 0:

(N+1) * 0`N`  +  (N+1) * 1`N`  +  (N+1) * 2`N` + ... (N+1) * x`N`   =   x`N + 1`

Sanity check with x=3, N = 2:

3(0)(1) + 3(1)(2) + 3(2)(3) + 3(3)(4) = (3)(4)(5)
0 + 6 + 18 + 36 = 60
PASSED

Divide both sides by N+1:

0`N` + 1`N` + 2`N` + 3`N` + ... + x`N` = x`N+1` / (N + 1)

Approximations are exactly correct at infinity:

(INF)`N` = (INF)^N

...and this is also exactly true once you get any substantial fraction of the way towards infinity.  Like, even (1,000,000)`3` starts to look pretty close to (1,000,000)^3.

As for all the finite numbers at the start of the sequence, they collectively comprise 0% of anything and can be ignored.

Conclusion:

0^N + 1^N + 2^N + 3^N + ... + INF^N = INF^(N+1) / (N+1).

So if you stack up an infinite number of squares with sides 0, 1, 2, 3... you will get exactly a third of an infinitely large cube, and more generally, if you stack up an infinite number of N-hypercubes with sides 0, 1, 2, 3... you'll get exactly 1 / (N+1) of an infinite (N+1)-hypercube.

Now, the nice thing about working with infinities is that you don't have to be very precise at all, and you'll still get exactly the right answer at the end!

Instead of taking x^N, let's use the 'rising powers', okay that didn't translate so probably is not a thing in Golarion, or at least the Share Language donor didn't know it.

Instead of taking x * x * x * ... * x, with N copies of x, let's consider the expression x * (x + 1) * (x + 2) ... until you've got N of those.

Let's denote that x`N`.

Obviously if x is any finite fraction of the way towards infinity, x`N` = x^N exactly.  Like, when you're working with trillions, a trillion squared is nearly the same number as a trillion times a trillion plus one, proportionally speaking; that's approximately true, so with infinities it's exactly true.

Example with x=10, N=3:

10`3`
= 10*11*12
= (10)(10+1)...(10+(3-1))    (3 terms total)
= (x)(x+1)...(x+(N-1))          (x=10, N=3, N terms total)

Now observe that:

11`3` - 10`3`
= 11*12*13 - 10*11*12
= (13 - 10) * (11 * 12)
= 3 * 11`2`

By pointwise generalization from 10 to x and 3 to N:

   (x + 1)`N`  -  (x)`N`
= (x + 1)(x + 2)...(x + N)  -  (x)(x+1)...(x + N - 1)
= ((x + N) - x) * (x+1)...(x + N - 1)
= N * (x+1)`N - 1`

Rearrange:

x`N` + N*(x+1)`N-1` = (x+1)`N`

Eliminate all subtraction whenever possible, in case something ends up 0 or negative that didn't start out that way:

x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

Sanity check with x=10, N=2:

10`2+1` + (2+1)*(10+1)`2` = (10+1)`2+1`
=> 10*11*12 + 3*(11*12) = 11*12*13
PASSED

Induction, starting from N * 0`N` = 0`N + 1` = 0:

(N+1) * 0`N`  +  (N+1) * 1`N`  +  (N+1) * 2`N` + ... (N+1) * x`N`   =   x`N + 1`

Sanity check with x=3, N = 2:

3(0)(1) + 3(1)(2) + 3(2)(3) + 3(3)(4) = (3)(4)(5)
0 + 6 + 18 + 36 = 60
PASSED

Divide both sides by N+1:

0`N` + 1`N` + 2`N` + 3`N` + ... + x`N` = x`N+1` / (N + 1)

Approximations are exactly correct at infinity:

(INF)`N` = (INF)^N

...and this is also exactly true once you get any substantial fraction of the way towards infinity.  Like, even (1,000,000)`3` starts to look pretty close to (1,000,000)^3.

As for all the finite numbers at the start of the sequence, they collectively comprise 0% of anything and can be ignored.

Conclusion:

0^N + 1^N + 2^N + 3^N + ... + INF^N = INF^(N+1) / (N+1).

So if you stack up an infinite number of squares with sides 0, 1, 2, 3... you will get exactly a third of an infinitely large cube, and more generally, if you stack up an infinite number of N-hypercubes with sides 0, 1, 2, 3... you'll get exactly 1 / (N+1) of an infinite (N+1)-hypercube.

Now, the nice thing about working with infinities is that you don't have to be very precise at all, and you'll still get exactly the right answer at the end!

Instead of taking x^N, let's use the 'rising powers', okay that didn't translate so probably is not a thing in Golarion, or at least the Share Language donor didn't know it.

Instead of taking x * x * x * ... * x, with N copies of x, let's consider the expression x * (x + 1) * (x + 2) ... until you've got N of those.

Let's denote that x`N`.

Obviously if x is any finite fraction of the way towards infinity, x`N` = x^N exactly.  Like, when you're working with trillions, a trillion squared is nearly the same number as a trillion times a trillion plus one, proportionally speaking; that's approximately true, so with infinities it's exactly true.

Example with x=10, N=3:

10`3`
= 10*11*12
= (10)(10+1)...(10+(3-1))    (3 terms total)
= (x)(x+1)...(x+(N-1))          (x=10, N=3)

Now observe that:

11`3` - 10`3`
= 11*12*13 - 10*11*12
= (13 - 10) * (11 * 12)
= 3 * 11`2`

By pointwise generalization from 10 to x and 3 to N:

   (x + 1)`N`  -  (x)`N`
= (x + 1)(x + 2)...(x + N)  -  (x)(x+1)...(x + N - 1)
= ((x + N) - x) * (x+1)...(x + N - 1)
= N * (x+1)`N - 1`

Rearrange:

x`N` + N*(x+1)`N-1` = (x+1)`N`

Eliminate all subtraction whenever possible, in case something ends up 0 or negative that didn't start out that way:

x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

Sanity check with x=10, N=2:

10`2+1` + (2+1)*(10+1)`2` = (10+1)`2+1`
=> 10*11*12 + 3*(11*12) = 11*12*13
PASSED

Induction, starting from N * 0`N` = 0`N + 1` = 0:

(N+1) * 0`N`  +  (N+1) * 1`N`  +  (N+1) * 2`N` + ... (N+1) * x`N`   =   x`N + 1`

Sanity check with x=3, N = 2:

3(0)(1) + 3(1)(2) + 3(2)(3) + 3(3)(4) = (3)(4)(5)
0 + 6 + 18 + 36 = 60
PASSED

Divide both sides by N+1:

0`N` + 1`N` + 2`N` + 3`N` + ... + x`N` = x`N+1` / (N + 1)

Approximations are exactly correct at infinity:

(INF)`N` = (INF)^N

...and this is also exactly true once you get any substantial fraction of the way towards infinity.  Like, even (1,000,000)`3` starts to look pretty close to (1,000,000)^3.

As for all the finite numbers at the start of the sequence, they collectively comprise 0% of anything and can be ignored.

Conclusion:

0^N + 1^N + 2^N + 3^N + ... + INF^N = INF^(N+1) / (N+1).

So if you stack up an infinite number of squares with sides 0, 1, 2, 3... you will get exactly a third of an infinitely large cube, and more generally, if you stack up an infinite number of N-hypercubes with sides 0, 1, 2, 3... you'll get exactly 1 / (N+1) of an infinite (N+1)-hypercube.

Now, the nice thing about working with infinities is that you don't have to be very precise at all, and you'll still get exactly the right answer at the end!

Instead of taking x^N, let's use the 'rising powers', okay that didn't translate so probably is not a thing in Golarion, or at least the Share Language donor didn't know it.

Instead of taking x * x * x * ... * x, with N copies of x, let's consider the expression x * (x + 1) * (x + 2) ... until you've got N of those.

Let's denote that x`N`.

Obviously if x is any finite fraction of the way towards infinity, x`N` = x^N exactly.  Like, when you're working with trillions, a trillion squared is nearly the same number as a trillion times a trillion plus one, proportionally speaking; that's approximately true, so with infinities it's exactly true.

Example with x=10, N=3:

10`3`
= 10*11*12
= (10)(10+1)...(10+(3-1))    (3 terms total)
= (x)(x+1)...(x+(N-1))          (x`N` with x=10, N=3)

Now observe that:

11`3` - 10`3`
= 11*12*13 - 10*11*12
= (13 - 10) * (11 * 12)
= 3 * 11`2`

By pointwise generalization from 10 to x and 3 to N:

   (x + 1)`N`  -  (x)`N`
= (x + 1)(x + 2)...(x + N)  -  (x)(x+1)...(x + N - 1)
= ((x + N) - x) * (x+1)...(x + N - 1)
= N * (x+1)`N - 1`

Rearrange:

x`N` + N*(x+1)`N-1` = (x+1)`N`

Eliminate all subtraction whenever possible, in case something ends up 0 or negative that didn't start out that way:

x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

Sanity check with x=10, N=2:

10`2+1` + (2+1)*(10+1)`2` = (10+1)`2+1`
=> 10*11*12 + 3*(11*12) = 11*12*13
PASSED

Induction, starting from N * 0`N` = 0`N + 1` = 0:

(N+1) * 0`N`  +  (N+1) * 1`N`  +  (N+1) * 2`N` + ... (N+1) * x`N`   =   x`N + 1`

Sanity check with x=3, N = 2:

3(0)(1) + 3(1)(2) + 3(2)(3) + 3(3)(4) = (3)(4)(5)
0 + 6 + 18 + 36 = 60
PASSED

Divide both sides by N+1:

0`N` + 1`N` + 2`N` + 3`N` + ... + x`N` = x`N+1` / (N + 1)

Approximations are exactly correct at infinity:

(INF)`N` = (INF)^N

...and this is also exactly true once you get any substantial fraction of the way towards infinity.  Like, even (1,000,000)`3` starts to look pretty close to (1,000,000)^3.

As for all the finite numbers at the start of the sequence, they collectively comprise 0% of anything and can be ignored.

Conclusion:

0^N + 1^N + 2^N + 3^N + ... + INF^N = INF^(N+1) / (N+1).

So if you stack up an infinite number of squares with sides 0, 1, 2, 3... you will get exactly a third of an infinitely large cube, and more generally, if you stack up an infinite number of N-hypercubes with sides 0, 1, 2, 3... you'll get exactly 1 / (N+1) of an infinite (N+1)-hypercube.

Now, the nice thing about working with infinities is that you don't have to be very precise at all, and you'll still get exactly the right answer at the end!

Instead of taking x^N, let's use the 'rising powers', okay that didn't translate so probably is not a thing in Golarion, or at least the Share Language donor didn't know it.

Instead of taking x * x * x * ... * x, with N copies of x, let's consider the expression x * (x + 1) * (x + 2) ... until you've got N of those.

Let's denote that x`N`.

Obviously if x is any finite fraction of the way towards infinity, x`N` = x^N exactly.  Like, when you're working with trillions, a trillion squared is nearly the same number as a trillion times a trillion plus one, proportionally speaking; that's approximately true, so with infinities it's exactly true.

Example with x=10, N=3:

10`3`
= 10*11*12
= (10)(10+1)...(10+(3-1))    (3 terms total)
= (x)(x+1)...(x+(N-1))          (x`N` with x=10, N=3)

Now observe that:

11`3` - 10`3`
= 11*12*13 - 10*11*12
= (13 - 10) * (11 * 12)
= 3 * 11`2`

By pointwise generalization from 10 to x and 3 to N:

   (x + 1)`N`  -  (x)`N`
= (x + 1)(x + 2)...(x + N)  -  (x)(x+1)...(x + N - 1)
= ((x + N) - x) * (x+1)...(x + N - 1)
= N * (x+1)`N - 1`

Rearrange:

(x + 1)`N`  -  (x)`N`   =  N * (x+1)`N - 1`
    ==>
x`N`  +  N*(x+1)`N-1`   =   (x+1)`N`

Eliminate all subtraction whenever possible, in case something ends up 0 or negative that didn't start out that way:

x`N`  +  N*(x+1)`N-1`   =   (x+1)`N`
    ==>
x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

Sanity check with x=10, N=2:

        x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`
=>   10`2+1` + (2+1)*(10+1)`2` = (10+1)`2+1`
=>   10*11*12 + 3*(11*12) = 11*12*13
PASSED

Induction, starting from N * 0`N` = 0`N+1` = 0:

(N+1) * 0`N`  +  (N+1) * 1`N`  +  (N+1) * 2`N` + ... (N+1) * x`N`   =   x`N + 1`

Sanity check with x=3, N = 2:

3(0)(1) + 3(1)(2) + 3(2)(3) + 3(3)(4) = (3)(4)(5)
0 + 6 + 18 + 36 = 60
PASSED

Divide both sides by N+1:

0`N` + 1`N` + 2`N` + 3`N` + ... + x`N` = x`N+1` / (N + 1)

Approximations are exactly correct at infinity:

(INF)`N` = (INF)^N

...and this is also exactly true once you get any substantial fraction of the way towards infinity.  Like, even (1,000,000)`3` starts to look pretty close to (1,000,000)^3.

As for all the finite numbers at the start of the sequence, they collectively comprise 0% of anything and can be ignored.

Conclusion:

0^N + 1^N + 2^N + 3^N + ... + INF^N = INF^(N+1) / (N+1).

So if you stack up an infinite number of squares with sides 0, 1, 2, 3... you will get exactly a third of an infinitely large cube, and more generally, if you stack up an infinite number of N-hypercubes with sides 0, 1, 2, 3... you'll get exactly 1 / (N+1) of an infinite (N+1)-hypercube.

Now, the nice thing about working with infinities is that you don't have to be very precise at all, and you'll still get exactly the right answer at the end!

Instead of taking x^N, let's use the 'rising powers', okay that didn't translate so probably is not a thing in Golarion, or at least the Share Language donor didn't know it.

Instead of taking x * x * x * ... * x, with N copies of x, let's consider the expression x * (x + 1) * (x + 2) ... until you've got N of those.

Let's denote that x`N`.

Obviously if x is any finite fraction of the way towards infinity, x`N` = x^N exactly.  Like, when you're working with trillions, a trillion squared is nearly the same number as a trillion times a trillion plus one, proportionally speaking; that's approximately true, so with infinities it's exactly true.

Example with x=10, N=3:

10`3`
= 10*11*12
= (10)(10+1)...(10+(3-1))    (3 terms total)
= (x)(x+1)...(x+(N-1))          (x`N` with x=10, N=3)

Now observe that:

11`3` - 10`3`
= 11*12*13 - 10*11*12
= (13 - 10) * (11 * 12)
= 3 * 11`2`

By pointwise generalization from 10 to x and 3 to N:

   (x + 1)`N`  -  (x)`N`
= (x + 1)(x + 2)...(x + N)  -  (x)(x+1)...(x + N - 1)
= ((x + N) - x) * (x+1)...(x + N - 1)
= N * (x+1)`N - 1`

Rearrange:

(x + 1)`N`  -  (x)`N`   =  N * (x+1)`N - 1`
    ==>
x`N`  +  N*(x+1)`N-1`   =   (x+1)`N`

We now have an N-1 term in our rule, which would be bad if N were 0.  Eliminate all subtraction whenever possible!  In case something ends up 0 or negative that didn't start out that way!

In this case we'll just redescribe the number N as being some other number N plus one, so N-1 goes to "N" and N goes to "N+1".  We coulda done that originally, but it woulda been needlessly more symbols.

x`N`  +  N*(x+1)`N-1`   =   (x+1)`N`
    ==>
x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

Sanity check with x=10, N=2:

        x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`
=>   10`2+1` + (2+1)*(10+1)`2` = (10+1)`2+1`
=>   10*11*12 + 3*(11*12) = 11*12*13
PASSED

Induction, starting from N * 0`N` = 0`N+1` = 0:

0`N+1` = 0
x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`
    ==>
(N+1) * 1`N`  +  (N+1) * 2`N` + ... (N+1) * x`N`   =   x`N + 1`
  (can optionally add a starting term of (N+1)*0`N+1' since this is equal to zero)

Sanity check with x=3, N = 2:

3(0)(1) + 3(1)(2) + 3(2)(3) + 3(3)(4) = (3)(4)(5)
0 + 6 + 18 + 36 = 60
PASSED

Divide both sides by N+1:

0`N` + 1`N` + 2`N` + 3`N` + ... + x`N` = x`N+1` / (N + 1)

Approximations are exactly correct at infinity:

(INF)`N` = (INF)^N

...and this is also exactly true once you get any substantial fraction of the way towards infinity.  Like, even (1,000,000)`3` starts to look pretty close to (1,000,000)^3.

As for all the finite numbers at the start of the sequence, they collectively comprise 0% of anything and can be ignored.

Conclusion:

0^N + 1^N + 2^N + 3^N + ... + INF^N = INF^(N+1) / (N+1).

So if you stack up an infinite number of squares with sides 0, 1, 2, 3... you will get exactly a third of an infinitely large cube, and more generally, if you stack up an infinite number of N-hypercubes with sides 0, 1, 2, 3... you'll get exactly 1 / (N+1) of an infinite (N+1)-hypercube.

Now, the nice thing about working with infinities is that you don't have to be very precise at all, and you'll still get exactly the right answer at the end!

Instead of taking x^N, let's use the 'rising powers', okay that didn't translate so probably is not a thing in Golarion, or at least the Share Language donor didn't know it.

Instead of taking x * x * x * ... * x, with N copies of x, let's consider the expression x * (x + 1) * (x + 2) ... until you've got N of those.

Let's denote that x`N`.

Obviously if x is any finite fraction of the way towards infinity, x`N` = x^N exactly.  Like, when you're working with trillions, a trillion squared is nearly the same number as a trillion times a trillion plus one, proportionally speaking; that's approximately true, so with infinities it's exactly true.

Example with x=10, N=3:

10`3`
= 10*11*12
= (10)(10+1)...(10+(3-1))    (3 terms total)
= (x)(x+1)...(x+(N-1))          (x`N` with x=10, N=3)

Now observe that:

11`3` - 10`3`
= 11*12*13 - 10*11*12
= (13 - 10) * (11 * 12)
= 3 * 11`2`

By pointwise generalization from 10 to x and 3 to N:

   (x + 1)`N`  -  (x)`N`
= (x + 1)(x + 2)...(x + N)  -  (x)(x+1)...(x + N - 1)
= ((x + N) - x) * (x+1)...(x + N - 1)
= N * (x+1)`N - 1`

Rearrange:

(x + 1)`N`  -  (x)`N`   =  N * (x+1)`N - 1`
    ==>
x`N`  +  N*(x+1)`N-1`   =   (x+1)`N`

We now have an N-1 term in our rule, which would be bad if N were 0.  Eliminate all subtraction whenever possible!  In case something ends up 0 or negative that didn't start out that way!

In this case we'll just redescribe the number N as being some other number N plus one, so N-1 goes to "N" and N goes to "N+1".  We coulda done that originally, but it woulda been needlessly more symbols.

x`N`  +  N*(x+1)`N-1`   =   (x+1)`N`
    ==>
x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

Sanity check with x=10, N=2:

        x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`
=>   10`2+1` + (2+1)*(10+1)`2` = (10+1)`2+1`
=>   10*11*12 + 3*(11*12) = 11*12*13
PASSED

Induction on x:

0`N+1` = 0    (base case)

x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

    ==>

(N+1) * 1`N`  +  (N+1) * 2`N` + ... (N+1) * x`N`   =   x`N + 1`
  (can optionally add a starting term of (N+1)*0`N+1' since this always equals zero)

Sanity check with x=3, N = 2:

3(0)(1) + 3(1)(2) + 3(2)(3) + 3(3)(4) = (3)(4)(5)
0 + 6 + 18 + 36 = 60
PASSED

Divide both sides by N+1:

0`N` + 1`N` + 2`N` + 3`N` + ... + x`N` = x`N+1` / (N + 1)

Approximations are exactly correct at infinity:

(INF)`N` = (INF)^N

...and this is also exactly true once you get any substantial fraction of the way towards infinity.  Like, even (1,000,000)`3` starts to look pretty close to (1,000,000)^3.

As for all the finite numbers at the start of the sequence, they collectively comprise 0% of anything and can be ignored.

Conclusion:

0^N + 1^N + 2^N + 3^N + ... + INF^N = INF^(N+1) / (N+1).

So if you stack up an infinite number of squares with sides 0, 1, 2, 3... you will get exactly a third of an infinitely large cube, and more generally, if you stack up an infinite number of N-hypercubes with sides 0, 1, 2, 3... you'll get exactly 1 / (N+1) of an infinite (N+1)-hypercube.

Now, the nice thing about working with infinities is that you don't have to be very precise at all, and you'll still get exactly the right answer at the end!

Instead of taking x^N, let's use the 'rising powers', okay that didn't translate so probably is not a thing in Golarion, or at least the Share Language donor didn't know it.

Instead of taking x * x * x * ... * x, with N copies of x, let's consider the expression x * (x + 1) * (x + 2) ... until you've got N of those.

Let's denote that x`N`.

Obviously if x is any finite fraction of the way towards infinity, x`N` = x^N exactly.  Like, when you're working with trillions, a trillion squared is nearly the same number as a trillion times a trillion plus one, proportionally speaking; that's approximately true, so with infinities it's exactly true.

Example with x=10, N=3:

10`3`
= 10*11*12
= (10)(10+1)...(10+(3-1))    (3 terms total)
= (x)(x+1)...(x+(N-1))          (x`N` with x=10, N=3)

Now observe that:

11`3` - 10`3`
= 11*12*13 - 10*11*12
= (13 - 10) * (11 * 12)
= 3 * 11`2`

By pointwise generalization from 10 to x and 3 to N:

   (x + 1)`N`  -  (x)`N`
= (x + 1)(x + 2)...(x + N)  -  (x)(x+1)...(x + N - 1)
= ((x + N) - x) * (x+1)...(x + N - 1)
= N * (x+1)`N - 1`

Rearrange:

(x + 1)`N`  -  (x)`N`   =  N * (x+1)`N - 1`
    ==>
x`N`  +  N*(x+1)`N-1`   =   (x+1)`N`

We now have an N-1 term in our rule, which would be bad if N were 0.  Eliminate all subtraction whenever possible!  In case something ends up 0 or negative that didn't start out that way!

In this case we'll just redescribe the number N as being some other number N plus one, so N-1 goes to "N" and N goes to "N+1".  We coulda done that originally, but it woulda been needlessly more symbols.

x`N`  +  N*(x+1)`N-1`   =   (x+1)`N`
    ==>
x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

Sanity check with x=10, N=2:

        x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`
=>   10`2+1` + (2+1)*(10+1)`2` = (10+1)`2+1`
=>   10*11*12 + 3*(11*12) = 11*12*13
PASSED

Then, by induction:

0`N+1`  =  0    (base case)

(x+1)`N+1`  =  x`N+1` + (N+1)*(x+1)`N`

    ==>

(N+1) * 1`N`  +  (N+1) * 2`N` + ... (N+1) * x`N`   =   x`N + 1`
  (can optionally add a starting term of (N+1)*0`N+1' since this always equals zero)

Sanity check with x=3, N = 2:

3(0)(1) + 3(1)(2) + 3(2)(3) + 3(3)(4) = (3)(4)(5)
0 + 6 + 18 + 36 = 60
PASSED

Divide both sides by N+1:

0`N` + 1`N` + 2`N` + 3`N` + ... + x`N` = x`N+1` / (N + 1)

Approximations are exactly correct at infinity:

(INF)`N` = (INF)^N

...and this is also exactly true once you get any substantial fraction of the way towards infinity.  Like, even (1,000,000)`3` starts to look pretty close to (1,000,000)^3.

As for all the finite numbers at the start of the sequence, they collectively comprise 0% of anything and can be ignored.

Conclusion:

0^N + 1^N + 2^N + 3^N + ... + INF^N = INF^(N+1) / (N+1).

So if you stack up an infinite number of squares with sides 0, 1, 2, 3... you will get exactly a third of an infinitely large cube, and more generally, if you stack up an infinite number of N-hypercubes with sides 0, 1, 2, 3... you'll get exactly 1 / (N+1) of an infinite (N+1)-hypercube.

Now, the nice thing about working with infinities is that you don't have to be very precise at all, and you'll still get exactly the right answer at the end!

Instead of taking x^N, let's use the 'rising powers', okay that didn't translate so probably is not a thing in Golarion, or at least the Share Language donor didn't know it.

Instead of taking x * x * x * ... * x, with N copies of x, let's consider the expression x * (x + 1) * (x + 2) ... until you've got N of those.

Let's denote that x`N`.

Obviously if x is any finite fraction of the way towards infinity, x`N` = x^N exactly.  Like, when you're working with trillions, a trillion squared is nearly the same number as a trillion times a trillion plus one, proportionally speaking; that's approximately true, so with infinities it's exactly true.

Example with x=10, N=3:

10`3`
= 10*11*12
= (10)(10+1)...(10+(3-1))    (3 terms total)
= (x)(x+1)...(x+(N-1))          (x`N` with x=10, N=3)

Now observe that:

11`3` - 10`3`
= 11*12*13 - 10*11*12
= (13 - 10) * (11 * 12)
= 3 * 11`2`

By pointwise generalization from 10 to x and 3 to N:

   (x + 1)`N`  -  (x)`N`
= (x + 1)(x + 2)...(x + N)  -  (x)(x+1)...(x + N - 1)
= ((x + N) - x) * (x+1)...(x + N - 1)
= N * (x+1)`N - 1`

Rearrange:

(x + 1)`N`  -  (x)`N`   =  N * (x+1)`N - 1`
    ==>
x`N`  +  N*(x+1)`N-1`   =   (x+1)`N`

We now have an N-1 term in our rule, which would be bad if N were 0.  Eliminate all subtraction whenever possible!  In case something ends up 0 or negative that didn't start out that way!

In this case we'll just redescribe the number N as being some other number N plus one, so N-1 goes to "N" and N goes to "N+1".  We coulda done that originally, but it woulda been needlessly more symbols.

x`N`  +  N*(x+1)`N-1`   =   (x+1)`N`
    ==>
x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

Sanity check with x=10, N=2:

        x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`
=>   10`2+1` + (2+1)*(10+1)`2` = (10+1)`2+1`
=>   10*11*12 + 3*(11*12) = 11*12*13
PASSED

Then, by induction on f(x) = x`N+1`:

0`N+1`  =  0    (base case)

(x+1)`N+1`  =  x`N+1` + (N+1)*(x+1)`N`

    ==>

 x`N + 1`  =  (N+1) * 1`N`  +  (N+1) * 2`N` + ... (N+1) * x`N`
  (can optionally add a starting term of (N+1)*0`N+1' since this always equals zero)

Sanity check with x=3, N = 2:

3(0)(1) + 3(1)(2) + 3(2)(3) + 3(3)(4) = (3)(4)(5)
0 + 6 + 18 + 36 = 60
PASSED

Divide both sides by N+1:

0`N` + 1`N` + 2`N` + 3`N` + ... + x`N` = x`N+1` / (N + 1)

Approximations are exactly correct at infinity:

(INF)`N` = (INF)^N

...and this is also exactly true once you get any substantial fraction of the way towards infinity.  Like, even (1,000,000)`3` starts to look pretty close to (1,000,000)^3.

As for all the finite numbers at the start of the sequence, they collectively comprise 0% of anything and can be ignored.

Conclusion:

0^N + 1^N + 2^N + 3^N + ... + INF^N = INF^(N+1) / (N+1).

So if you stack up an infinite number of squares with sides 0, 1, 2, 3... you will get exactly a third of an infinitely large cube, and more generally, if you stack up an infinite number of N-hypercubes with sides 0, 1, 2, 3... you'll get exactly 1 / (N+1) of an infinite (N+1)-hypercube.

Now, the nice thing about working with infinities is that you don't have to be very precise at all, and you'll still get exactly the right answer at the end!

Instead of taking x^N, let's use the 'rising powers', okay that didn't translate so probably is not a thing in Golarion, or at least the Share Language donor didn't know it.

Instead of taking x * x * x * ... * x, with N copies of x, let's consider the expression x * (x + 1) * (x + 2) ... until you've got N of those.

Let's denote that x`N`.

Obviously if x is any finite fraction of the way towards infinity, x`N` = x^N exactly.  Like, when you're working with trillions, a trillion squared is nearly the same number as a trillion times a trillion plus one, proportionally speaking; that's approximately true, so with infinities it's exactly true.

Example with x=10, N=3:

10`3`
= 10*11*12
= (10)(10+1)...(10+(3-1))    (3 terms total)
= (x)(x+1)...(x+(N-1))          (x`N` with x=10, N=3)

Now observe that:

11`3` - 10`3`
= 11*12*13 - 10*11*12
= (13 - 10) * (11 * 12)
= 3 * 11`2`

By pointwise generalization from 10 to x and 3 to N:

   (x + 1)`N`  -  (x)`N`
= (x + 1)(x + 2)...(x + N)  -  (x)(x+1)...(x + N - 1)
= ((x + N) - x) * (x+1)...(x + N - 1)
= N * (x+1)`N - 1`

Rearrange:

(x + 1)`N`  -  (x)`N`   =  N * (x+1)`N - 1`
    ==>
x`N`  +  N*(x+1)`N-1`   =   (x+1)`N`

We now have an N-1 term in our rule, which would be bad if N were 0.  Eliminate all subtraction whenever possible!  In case something ends up 0 or negative that didn't start out that way!

In this case we'll just redescribe the number N as being some other number N plus one, so N-1 goes to "N" and N goes to "N+1".  We coulda done that originally, but it woulda been needlessly more symbols.

x`N`  +  N*(x+1)`N-1`   =   (x+1)`N`
    ==>
x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

Sanity check with x=10, N=2:

        x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`
=>   10`2+1` + (2+1)*(10+1)`2` = (10+1)`2+1`
=>   10*11*12 + 3*(11*12) = 11*12*13
PASSED

Then, by induction on f(x) = x`N+1`:

0`N+1`  =  0    (base case)

(x+1)`N+1`  =  x`N+1` + (N+1)*(x+1)`N`

    ==>

 x`N + 1`  =  (N+1) * 1`N`  +  (N+1) * 2`N` + ... (N+1) * x`N`
  (can optionally add a starting term of (N+1)*0`N+1' since this always equals zero)

Sanity check with x=3, N = 2:

(3)(4)(5)  =  3(0)(1) + 3(1)(2) + 3(2)(3) + 3(3)(4)
60  =  0 + 6 + 18 + 36
PASSED

Divide both sides by N+1:

0`N` + 1`N` + 2`N` + 3`N` + ... + x`N` = x`N+1` / (N + 1)

Approximations are exactly correct at infinity:

(INF)`N` = (INF)^N

...and this is also exactly true once you get any substantial fraction of the way towards infinity.  Like, even (1,000,000)`3` starts to look pretty close to (1,000,000)^3.

As for all the finite numbers at the start of the sequence, they collectively comprise 0% of anything and can be ignored.

Conclusion:

0^N + 1^N + 2^N + 3^N + ... + INF^N = INF^(N+1) / (N+1).

So if you stack up an infinite number of squares with sides 0, 1, 2, 3... you will get exactly a third of an infinitely large cube, and more generally, if you stack up an infinite number of N-hypercubes with sides 0, 1, 2, 3... you'll get exactly 1 / (N+1) of an infinite (N+1)-hypercube.

Now, the nice thing about working with infinities is that you don't have to be very precise at all, and you'll still get exactly the right answer at the end!

Instead of taking x^N, let's use the 'rising powers', okay that didn't translate so probably is not a thing in Golarion, or at least the Share Language donor didn't know it.

Instead of taking x * x * x * ... * x, with N copies of x, let's consider the expression x * (x + 1) * (x + 2) ... until you've got N of those.

Let's denote that x`N`.

Obviously if x is any finite fraction of the way towards infinity, x`N` = x^N exactly.  Like, when you're working with trillions, a trillion squared is nearly the same number as a trillion times a trillion plus one, proportionally speaking; that's approximately true, so with infinities it's exactly true.

Example with x=10, N=3:

10`3`
= 10*11*12
= (10)(10+1)...(10+(3-1))    (3 terms total)
= (x)(x+1)...(x+(N-1))          (x`N` with x=10, N=3)

Now observe that:

11`3` - 10`3`
= 11*12*13 - 10*11*12
= (13 - 10) * (11 * 12)
= 3 * 11`2`

By pointwise generalization from 10 to x and 3 to N:

   (x + 1)`N`  -  (x)`N`
= (x + 1)(x + 2)...(x + N)  -  (x)(x+1)...(x + N - 1)
= ((x + N) - x) * (x+1)...(x + N - 1)
= N * (x+1)`N - 1`

Rearrange:

(x + 1)`N`  -  (x)`N`   =  N * (x+1)`N - 1`
    ==>
x`N`  +  N*(x+1)`N-1`   =   (x+1)`N`

We now have an N-1 term in our rule, which would be bad if N were 0.  Eliminate all subtraction whenever possible!  In case something ends up 0 or negative that didn't start out that way!

In this case we'll just redescribe the number N as being some other number N plus one, so N-1 goes to "N" and N goes to "N+1".  We coulda done that originally, but it woulda been needlessly more symbols.

x`N`  +  N*(x+1)`N-1`   =   (x+1)`N`
    ==>
x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`

Sanity check with x=10, N=2:

        x`N+1` + (N+1)*(x+1)`N` = (x+1)`N+1`
=>   10`2+1` + (2+1)*(10+1)`2` = (10+1)`2+1`
=>   10*11*12 + 3*(11*12) = 11*12*13
PASSED

Then, by induction on f(x) = x`N+1`:

0`N+1`  =  0    (base case)

(x+1)`N+1`  =  x`N+1` + (N+1)*(x+1)`N`

    ==>

 x`N + 1`  =  (N+1) * 1`N`  +  (N+1) * 2`N` + ... (N+1) * x`N`
  (can optionally add a starting term of (N+1)*0`N+1' since this always equals zero)

Sanity check with x=3, N = 2:

(3)(4)(5)  =  3(0)(1) + 3(1)(2) + 3(2)(3) + 3(3)(4)
60  =  0 + 6 + 18 + 36
PASSED

Divide both sides by N+1:

x`N+1` / (N + 1)  =  1`N` + 2`N` + 3`N` + ... + x`N`

Approximations are exactly correct at infinity:

(INF)`N` = (INF)^N

...and this is also exactly true once you get any substantial fraction of the way towards infinity.  Like, even (1,000,000)`3` starts to look pretty close to (1,000,000)^3.

As for all the finite numbers at the start of the sequence, where the equality isn't exact, they collectively comprise 0% of anything and can be ignored.

Conclusion:

0^N + 1^N + 2^N + 3^N + ... + INF^N = INF^(N+1) / (N+1).

So if you stack up an infinite number of squares with sides 0, 1, 2, 3... you will get exactly a third of an infinitely large cube, and more generally, if you stack up an infinite number of N-hypercubes with sides 0, 1, 2, 3... you'll get exactly 1 / (N+1) of an infinite (N+1)-hypercube.