Show $\int_0^1 x^n (1-x)^{N-n} dx = \left(\binom{N}{n} (N+1)\right)^{-1}$

Question

As the title suggests I would like to know how to solve the following integral (solution found w Mathematica)

$$\int_0^1 x^n (1-x)^{N-n} dx = \left(\binom{N}{n} (N+1)\right)^{-1}$$ (we may assume $0<n<N$ and $n,N\in \mathbb{N}$)

I'm thinking there's probably a cool trick using the binomial theorem but I can't figure it out.

Thanks

Edit work in progress

\begin{align} \int_0^1 x^{n} (1-x)^{N-{n}} dx &= \int_0^1 \sum _{k=0}^{N-n} \binom{N-n}{k} (-1)^k x^{n+k} dx\\ &=\sum _{k=0}^{N-n} \binom{N-n}{k}(-1)^k (1+k+n)^{-1}\\ &\ldots \end{align} Now I'm thinking to multiply by $\binom{N}{n}$ and show that sum simply reduces to $(N+1)^{-1}$

Answer 1

An example of trying to have simple polynomial identities do the combinatorial work is as follows: observe that the integral is $1/{N\choose n}$ times the coefficient of $t^{N-n}$ in $\int_0^1 (x+t(1-x))^N\,dx$. The latter integral is readily computed to be ${1\over N+1}{1-t^{N+1}\over 1-t}$, which is ${1\over N+1}(1+t+t^2+\ldots+t^N)$. :)

EDIT: expanding $(x+t(1-x))^N$ by the binomial theorem: $\sum_{n=0}^N {N\choose n} x^n(1-x)^{N-n}\cdot t^{N-n}$.