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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)$$\int{x^n(1-x)^m}{dx} = \int{x^n\cdot\left(\sum^m_{k=0}\binom{m}{k}(-1)^kx^k\right)}{dx} = \int{\left(\sum^m_{k=0}\binom{m}{k}(-1)^kx^kx^n\right)}{dx} = $$ $$ = \sum^m_{k=0}\left(\binom{m}{k}(-1)^k\cdot\int{x^{n+k}}{dx}\right) = \sum^m_{k=0}\binom{m}{k}(-1)^k\frac{x^{n+k+1}}{n+k+1} =: F(x).$$ $$F(1) - F(0) = \sum^m_{k=0}\frac{\binom{m}{k}(-1)^k1^{n+k}}{n+k+1} - \sum^m_{k=0}\frac{\binom{m}{k}(-1)^k0^{n+k}}{n+k+1} = \sum^m_{k=0}\frac{\binom{m}{k}(-1)^k}{n+k+1}$$

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    $\begingroup$ Everything is fine, but $$\int_{0}^{1}x^n(1-x)^m\,dx = \frac{n!m!}{(n+m+1)!}$$ directly follows from Euler's Beta function. $\endgroup$ Commented Jan 7, 2018 at 20:01

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Induction on $m$.

The $m=0$ case is trivial.

For $m>0$,

$$ \begin{align} \sum_{k=0}^m \frac{\binom{m}{k}(-1)^k}{n+k+1} &= \sum_{k=1}^m \frac{\binom{m-1}{k-1}(-1)^k}{n+k+1} + \sum_{k=0}^{m-1} \frac{\binom{m-1}{k}(-1)^k}{n+k+1} \\ &= \sum_{j=0}^{m-1} \frac{\binom{m-1}{j}(-1)^{j+1}}{(n+1)+j+1} + \sum_{k=0}^{m-1} \frac{\binom{m-1}{k}(-1)^k}{n+k+1} \\ &= \frac{-(m-1)!(n+1)!}{(m+n+1)!} + \frac{(m-1)!n!}{(m+n)!} \\ &= (-(n+1)+(m+n+1))\frac{(m-1)!n!}{(m+n+1)!} \\ &= \frac{m!n!}{(m+n+1)!} \end{align} $$

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Here is a more direct approach along the lines you were perhaps thinking.

Notice that $$\int_0^1 x^{n + k} \, dx = \frac{1}{n + k + 1}.$$ With this result your sum can be rewritten as $$S = \sum_{k = 0}^m \binom{m}{k} (-1)^k \int_0^1 x^{n + k} \, dx = \int_0^1 x^n \left [\sum_{k = 0}^m \binom{m}{k} (-1)^k x^k \right ] \, dx.$$

The sum appearing within the square brackets is nothing more than the binomial sum $$(1 - x)^m = \sum_{k = 0}^m \binom{m}{k} (-1)^k x^k.$$ Thus \begin{align*} S &= \int_0^1 x^n (1 - x)^m \, dx\\ &= \int_0^1 x^{(n + 1) - 1} (1 - x)^{(m + 1) - 1} \, dx\\ &= \text{B}(n + 1, m + 1)\\ &= \frac{\Gamma (n + 1) \Gamma (m + 1)}{\Gamma (n + m + 2)}\\ &= \frac{n! m!}{(n + m + 1)!}, \end{align*} as required. Here, as noted by Jack, $\text{B}(x,y)$ corresponds to Euler's Beta function.

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