<s>[INST] Prove that $\sum^m_{k=0}\binom{m}{k}(-1)^k(\frac{1}{n+k+1}) = \frac{m!n!}{(m+n+1)!}$ I am looking for a more direct proof of the following identity: $$\sum^m_{k=0}\frac{\binom{m}{k}(-1)^k}{n+k+1} = \frac{m!n!}{(m+n+1)!}$$
My original proof comes from evaluating $\int^1_0{x^n(1-x)^m}{dx}$, $n, m \in \mathbb{N}_0$ in two ways:
1) Let $u = x^n$, $u' = nx^{n-1}$, $v' = (1-x)^m$, $v = -\frac{(1-x)^{m+1}}{m+1}$:
$$\int^1_0{x^n(1-x)^m}{dx} = \frac{x^n(1-x)^{m+1}}{m+1} + \frac{n}{m+1}\int^1_0{x^{n-1}(1-x)^{m+1}}{dx}$$
$x=1\Rightarrow \frac{x^n(1-x)^{m+1}}{m+1} = 0$ and $x=0\Rightarrow \frac{x^n(1-x)^{m+1}}{m+1} = 0$, therefore
$$\int^1_0{x^n(1-x)^m}{dx} = \frac{n}{m+1}\int^1_0{x^{n-1}(1-x)^{m+1}}{dx}$$
Repeating the process $n$ times gives: 
$$\int^1_0{x^n(1-x)^m}{dx} = \frac{m!n!}{(m+n)!}\int^1_0{x^0(1-x)^{m+n}}{dx} = $$$$=\frac{m!n!}{(m+n)!}\cdot\frac{(1-1)^{m+n+1}}{m+n+1} - \frac{m!n!}{(m+n)!}\cdot\left(-\frac{(1-0)^{m+n+1}}{m+n+1}\right) = \frac{m!n!}{(m+n+1)!}$$
2)$$\int{x^n(1-x)^m}{dx} = \int{x^n\cdot\left(\sum^m_{k=0}\binom{m}{k}(-1)^kx^k\right)}{dx} = \int{\left(\sum^m_{k=0}\binom{m}{k}(-1)^kx^kx^n\right)}{dx} = $$
$$ = \sum^m_{k=0}\left(\binom{m}{k}(-1)^k\cdot\int{x^{n+k}}{dx}\right) = \sum^m_{k=0}\binom{m}{k}(-1)^k\frac{x^{n+k+1}}{n+k+1} =: F(x).$$
$$F(1) - F(0) = \sum^m_{k=0}\frac{\binom{m}{k}(-1)^k1^{n+k}}{n+k+1} - \sum^m_{k=0}\frac{\binom{m}{k}(-1)^k0^{n+k}}{n+k+1} = \sum^m_{k=0}\frac{\binom{m}{k}(-1)^k}{n+k+1}$$
 [/INST] Induction on $m$.
The $m=0$ case is trivial.
For $m>0$,
$$
\begin{align}
\sum_{k=0}^m \frac{\binom{m}{k}(-1)^k}{n+k+1} &= 
\sum_{k=1}^m \frac{\binom{m-1}{k-1}(-1)^k}{n+k+1} + \sum_{k=0}^{m-1} \frac{\binom{m-1}{k}(-1)^k}{n+k+1} \\ &=
\sum_{j=0}^{m-1} \frac{\binom{m-1}{j}(-1)^{j+1}}{(n+1)+j+1} + \sum_{k=0}^{m-1} \frac{\binom{m-1}{k}(-1)^k}{n+k+1} \\ &=
\frac{-(m-1)!(n+1)!}{(m+n+1)!} + \frac{(m-1)!n!}{(m+n)!} \\ &=
(-(n+1)+(m+n+1))\frac{(m-1)!n!}{(m+n+1)!} \\ &=
\frac{m!n!}{(m+n+1)!}
\end{align}
$$
</s>