<s>[INST] 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.
 [/INST] 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.
</s>