<s>[INST] Prove the formula of combination number $$
\sum_{k=0}^n{\left( \begin{array}{c}
 n\\
 k\\
\end{array} \right) \frac{\left( -1 \right) ^k}{x+k}=\frac{n!}{x\left( x+1 \right) \cdots \left( x+n \right)}}
\\
$$
I recently read an article and found this formula above. I tried to prove it, but only proved the following result. I don't know how to prove this more general combinatorial equation.
$$
\sum_{k=0}^n{\left( \begin{array}{c}
 n\\
 k\\
\end{array} \right) \frac{\left( -1 \right) ^k}{m+k+1}=\int_0^1{\sum_{k=0}^n{\left( \begin{array}{c}
 n\\
 k\\
\end{array} \right) \left( -1 \right) ^kx^{m+k}\text{d}x}=\int_0^1{x^m\left( 1-x \right) ^n}\text{d}x}=\frac{m!n!}{\left( m+n+1 \right) !}}
$$
 [/INST] By partial fractions, we know
$$\frac{n!}{x(x+1) \cdots (x+n)} = \frac{a_0}{x} + \frac{a_1}{x+1} + \dots + \frac{a_n}{x+n}$$
for some set of numbers $a_0, a_1, \dots , a_n$. To find $a_k$ we multiply both sides of the equation by $x+k$ and then set $x=-k$ in the resulting equation.  The result is
$$a_k = (-1)^k \binom{n}{k}$$
</s>