Show that$ \sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{k+n+1}=\frac{n!\,m!}{(n+m+1)!}$ 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.
 A: 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}
A: In trying to prove
$$S_m = \sum_{k=0}^m \frac{(-1)^k}{k+n+1} {m\choose k}
= \frac{n! m!}{(n+m+1)!}$$
we introduce
$$f(z) = m! (-1)^m \frac{1}{z+n+1} \prod_{q=0}^m \frac{1}{z-q}.$$
This has the property that for $0\le k\le m$
$$\mathrm{Res}_{z=k} f(z)
= m! (-1)^m \frac{1}{k+n+1}
\prod_{q=0}^{k-1} \frac{1}{k-q}
\prod_{q=k+1}^m \frac{1}{k-q}
\\ = m! (-1)^m \frac{1}{k+n+1}
\frac{1}{k!} \frac{(-1)^{m-k}}{(m-k)!}
= \frac{(-1)^k}{k+n+1} {m\choose k}.$$
Note that the residue of $f(z)$ at infinity is zero by inspection and
residues sum to zero so that we have
$$S_m = \sum_{k=0}^m \mathrm{Res}_{z=k} f(z)
= - \mathrm{Res}_{z=-n-1} f(z)
\\ = - \mathrm{Res}_{z=-n-1} \; m! (-1)^m
\frac{1}{z+n+1} \prod_{q=0}^m \frac{1}{z-q}
= - m! (-1)^m
\prod_{q=0}^m \frac{1}{-n-1-q}
\\ = m! \prod_{q=0}^m \frac{1}{n+1+q}
= \frac{m! n!}{(n+m+1)!}.$$
This is the claim.
A: There is the “Linear algebra” way. Let $E$ be the space of functions from $(0,\infty)$ to $\Bbb R$ and define the operator $S\colon f\mapsto Sf$ on $E$ by $$Sf(x):=f(x+1),\quad f\in E,\ x>0.$$
Let the difference operator $\Delta:=\mathrm{Id}_E-S$. Since
$\mathrm{Id}_E$ and $-S$ commute, the Binomial theorem gives you $\Delta^m$ for any integer $m\ge0$:
$$\Delta^m=\sum_{k=0}^m\binom mk(\mathrm{Id}_E)^{m-k}(-S)^k,$$
that is
$$\Delta^mf(x)=\sum_{k=0}^m\binom mk(-1)^kf(x+k).\tag{$\star$}$$
Now consider $f\in E$ given by $f(x):=\frac1x$. We can see that
\begin{align*}
\Delta f(x)&=\frac1x-\frac1{x+1}=\frac1{x(x+1)},\\[.4em]
\Delta^2f(x):=\Delta(\Delta f)(x)&=\frac1{x(x+1)}-\frac1{(x+1)(x+2)}=\frac2{x(x+1)(x+2)},
\end{align*}
and an immediate induction shows that
$$\Delta^m f(x)=\frac{m!}{x(x+1)\cdots(x+m)}.$$
Thus $(\star)$ reads
$$\frac{m!}{x(x+1)\cdots(x+m)}=\sum_{k=0}^m\binom mk(-1)^k\frac1{x+k}.$$
This is an identity between functions of the variable $x>0$ and it suffices to take $x:=n+1$ to establish your claim.
