Show that$ \sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{k+n+1}=\frac{n!\,m!}{(n+m+1)!}$

Question

I want to show that
\begin{align}
&\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{k+n+1}=\frac{n!\,m!}{(n+m+1)!}\\
\end{align}

I came across this when trying to prove
\begin{align} \int_0^1 (1-x)^m x^n dx =\frac{n!\,m!}{(n+m+1)!}\\
\end{align}
My teacher proved it by using the substitution $x=\sin^2 (t)$ and then using the Wallis formula but I tried to prove it by using Binomial Theorem and integrating each term and I got stuck at this step.

Answer 1

Let $k\in\mathbb{N} $, we have the following : \begin{aligned}I_{n+k,m-k}=\int_{0}^{1}{x^{n+k}\left(1-x\right)^{m-k}\,\mathrm{d}x}&=\left[\frac{x^{n+k+1}\left(1-x\right)^{m-k}}{n+k+1}\right]_{0}^{1}+\frac{m-k}{n+k+1}\int_{0}^{1}{x^{n+k+1}\left(1-x\right)^{m-k-1}\,\mathrm{d}x}\\ I_{n+k,m-k}&=\frac{m-k}{n+k+1}I_{n+k+1,m-k-1}\\ \Longrightarrow\prod_{k=0}^{m-1}{\frac{I_{n+k,m-k}}{I_{n+k+1,m-k-1}}}&=\prod_{k=0}^{m-1}{\frac{m-k}{n+k+1}}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{I_{n,m}}{I_{n+m,0}}&=\prod_{k=0}^{m-1}{\frac{m-k}{n+k+1}}\\ I_{n,m}&=\frac{n!m!}{\left(n+m+1\right)!}\end{aligned}