Identity of $\frac{n!}{x(x+1)(x+2)...(x+n)}$ 
$$\frac{n!}{x(x+1)(x+2)\dots(x+n)}=\sum_{k=0}^n{n\choose k}\frac{(-1)^k}{x+k}$$

This is the identity. I tried proving it by induction but it got really complicated and couldn't solve it.
 A: A proof by induction is possible, using the identity $$\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}.$$  To this end, let $$P_n(x) = \sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{x+k}.$$  Then $$\begin{align*} P_{n+1}(x) &= \sum_{k=0}^{n+1} \left(\binom{n}{k-1} + \binom{n}{k}\right) \frac{(-1)^k}{x+k} \\ 
&= \sum_{m=0}^n \binom{n}{m} \frac{(-1)^{m+1}}{x+1+m} + \sum_{m=0}^n \binom{n}{m} \frac{(-1)^m}{x+m} \\
&= -P_n (x+1) + P_n (x).
\end{align*}$$
Now let $Q_n(x) = \frac{n!}{x(x+1)\ldots(x+n)}$.  Then if $P_n(x) = Q_n(x)$ for some positive integer $n$, we have 
\begin{align}
P_{n+1}(x) &= -Q_n(x+1) + Q_n(x) \\
 &= -\frac{n!}{(x+1)(x+2)\ldots(x+n+1)} + \frac{n!}{x(x+1)\ldots(x+n)} \\
 &= \frac{n!}{(x+1)(x+2)\ldots(x+n)} \left( \frac{1}{x} - \frac{1}{x+n+1} \right) \\
&= \frac{n! (n+1)}{x(x+1)\ldots(x+n+1)} \\
&= Q_{n+1}(x),
\end{align} completing the induction step.
A: Not an inductive proof but here you are: 
Separate the following into partial fractions:
$$\frac{1}{(x+1)(x+2)\cdots (x+n)}\equiv \sum_{k=0}^{n}\frac{A_k}{x+k}$$
where the $A_k$ are to be determined.
Multiplying through by $x(x+1)(x+2)\cdots (x+n)$
$$1\equiv \sum_{k=0}^{n}A_k \prod_{\begin{subarray}{c} 0 \le i \le n\\ i\ne k\end{subarray} }(x+i)$$
Then setting $x=-k$ yields
$$1=A_k\cdot -k\cdot -(k-1)\:\cdots\: -2\cdot -1\cdot 1\cdot 2\:\cdots\: (n-k-1)\cdot (n-k)$$
$$1 = A_k(-1)^kk!(n-k)!$$
$$\implies\: A_k=\frac{(-1)^k}{k!(n-k)!}$$
hence
$$\begin{align}\frac{n!}{(x+1)(x+2)\cdots (x+n)}&=\sum_{k=0}^{n}(-1)^k\frac{n!}{k!(n-k)!}\frac{1}{x+k}\\&= \sum_{k=0}^{n}(-1)^k\binom{n}{k}\frac{1}{x+k}\end{align}$$
A: 
$$\sum_{k=0}^n{n\choose k}\frac{(-1)^k}{x+k}$$

Note that ,
$$\int_{0}^{1} y^{x+k-1} dy=\frac{1}{x+k}$$
Also note that $y^{x+k-1}=y^{x-1} y^{k}$. So we have,

$$\sum_{k=0}^n{n\choose k}\frac{(-1)^k}{x+k}$$
$$=\int_{0}^{1} \sum_{k=0}^n{n\choose k}(-1)^ky^{x-1} y^{k} dy$$
$$=\int_{0}^{1} y^{x-1} \sum_{k \geq 0} {n \choose k} (-y)^k dy$$
$$=\int_{0}^{1} y^{x-1}(1-y)^n dy$$
$$=B(x,n+1)$$

Now by the Beta-Gamma function relationship this is,
$$=\frac{\Gamma (x)\Gamma (n+1)}{\Gamma (x+n+1)}$$
Now given a nonnegative integer $n$ so the original summand makes sense, by the property $\Gamma(x+1)=x\Gamma(x)$ we have,
$$\begin{align*} \Gamma(x+n+1)=\Gamma (x) \left( (x)(x+1)...(x+n) \right) \end{align*}$$
So that,
$$\begin{align} \frac{\Gamma(x+n+1)}{\Gamma(x)}=(x)(x+1)...(x+n) \end{align}$$
$$\frac{\Gamma (x)\Gamma (n+1)}{\Gamma (x+n+1)}=\frac{n!}{(x)(x+1)...(x+n)}$$
