If you're willing to assume your audience can already freely wield the machinery of derivatives and antiderivatives and understand on an intuitive level everything that they mean, then sure, you can solve the problem more quickly using overpowered tools like that.

Integrating x^N gives you x^(N+1)/(N+1), which evaluated from 0 to 1 will give you 1/(N+1), yes.

Given a mix of N LEFTs and M RIGHTs, however, and the need to renormalize the posterior over that, they might find themselves in a bit more trouble when it comes to deriving the integral...

Integral of p from 0 to 1:  [  (1 - p)^M p^N dp  ]   =   M! * N! / (M + N + 1)!

If you're willing to assume your audience can already freely wield the machinery of derivatives and antiderivatives and understand on an intuitive level everything that they mean, then sure, you can solve the problem more quickly using overpowered tools like that.

Integrating x^N gives you x^(N+1)/(N+1), which evaluated from 0 to 1 will give you 1/(N+1), yes.

Given a mix of N LEFTs and M RIGHTs, however, and the need to renormalize the posterior over that, they might find themselves in a bit more trouble when it comes to deriving the integral...

Integral of p from 0 to 1:  [  (1 - p)^M p^N dp  ]   =   M! * N! / (M + N + 1)!

If you're willing to assume your audience can already freely wield the machinery of derivatives and antiderivatives and understand on an intuitive level everything that they mean, then sure, you can solve the problem more quickly using overpowered tools like that.

Integrating x^N gives you x^(N+1)/(N+1), which evaluated from 0 to 1 will give you 1/(N+1), yes.

Given a mix of N LEFTs and M RIGHTs, however, and the need to renormalize the posterior over that, they might find themselves in a bit more trouble when it comes to deriving the integral...

Integral of p from 0 to 1:  [  (1 - p)^M p^N dp  ]   =   M! * N! / (M + N + 1)!

If you're willing to assume your audience can already freely wield the machinery of derivatives and antiderivatives and understand on an intuitive level everything that they mean, then sure, you can solve the problem more quickly using overpowered tools like that.

Integrating x^N gives you x^(N+1)/(N+1), which evaluated from 0 to 1 will give you 1/(N+1).

Given a mix of N LEFTs and M RIGHTs, however, and the need to renormalize the posterior over that, they might find themselves in a bit more trouble when it comes to deriving the integral...

Integral of p from 0 to 1:  [  (1 - p)^M p^N dp  ]   =   M! * N! / (M + N + 1)!