A proof of the identity $ \sum_{k = 0}^{n} \frac{(-1)^{k} \binom{n}{k}}{x + k} = \frac{n!}{(x + 0) (x + 1) \cdots (x + n)} $. I have to prove that
$$
\forall n \in \mathbb{N}_{0}, ~ \forall x \in \mathbb{R} \setminus \mathbb{N}_{0}:
\qquad
  \sum_{k = 0}^{n} \frac{(-1)^{k} \binom{n}{k}}{x + k}
= \frac{n!}{(x + 0) (x + 1) \cdots (x + n)}.
$$
However, I was unable to find a proof. I have tried to use the binomial expansion of $ (1 + x)^{n} $ to get the l.h.s., by performing a suitable multiplication followed by integration, but I was unable to obtain the required form. Please help me out with the proof. Thanks in advance.

Remark. What follows  is an exact duplicate of  an earlier post of
mine  which  I am  unable  to  locate  despite making  a  considerable
effort. 
We seek to verify that
$$\sum_{k=0}^n \frac{1}{x+k} (-1)^k {n\choose k}
= n! \times \prod_{q=0}^n \frac{1}{x+q}.$$
Consider the function
$$f(z) = n! \frac{1}{x+z} \prod_{q=0}^n \frac{1}{z-q}$$
where clearly  $x$ is not equal  to $0,-1,-2,\ldots -n$ or  the LHS of
the target formula becomes singular.
We compute  the residues of $f(z).$  We get for the  poles at $p\in
[0,n]$ the residue
$$\mathrm{Res}_{z=p} f(z) =
n!\frac{1}{x+p} \prod_{q=0}^{p-1} \frac{1}{p-q}
\prod_{q=p+1}^n \frac{1}{p-q}
\\ = n! \frac{1}{x+p} \frac{1}{p!} \frac{(-1)^{n-p}}{(n-p)!}
= \frac{1}{x+p} (-1)^{n-p} {n\choose p}.$$
The residue at $z=-x$ yields
$$\mathrm{Res}_{z=-x} f(z) =
n! \prod_{q=0}^n \frac{1}{-x-q}
= n! (-1)^{n+1} \times
\prod_{q=0}^n \frac{1}{x+q}.$$
The residue at infinity is
$$\mathrm{Res}_{z=\infty} f(z) =
- \mathrm{Res}_{z=0} \frac{1}{z^2} f(1/z)
\\ = - \mathrm{Res}_{z=0} \frac{1}{z^2}
n! \frac{1}{x+1/z} \prod_{q=0}^n \frac{1}{1/z-q}
\\ = - \mathrm{Res}_{z=0} \frac{1}{z^2}
n! \frac{z}{zx+1} \prod_{q=0}^n \frac{z}{1-qz}
\\ = - \mathrm{Res}_{z=0} \frac{z^{n+2}}{z^2}
n! \frac{1}{zx+1} \prod_{q=0}^n \frac{1}{1-qz}
\\ = - \mathrm{Res}_{z=0} z^n
n! \frac{1}{zx+1} \prod_{q=0}^n \frac{1}{1-qz}
= 0.$$
To  conclude observe  that the  residues  sum to  zero and  collecting
everything we obtain
$$\sum_{p=0}^n\frac{1}{x+p} (-1)^{n-p} {n\choose p}
+ n! (-1)^{n+1} \times
\prod_{q=0}^n \frac{1}{x+q} = 0.$$
This is
$$\sum_{p=0}^n\frac{1}{x+p} (-1)^{p} {n\choose p}
- n! \times
\prod_{q=0}^n \frac{1}{x+q} = 0.$$
or
$$\sum_{p=0}^n\frac{1}{x+p} (-1)^{p} {n\choose p}
= n! \times
\prod_{q=0}^n \frac{1}{x+q}$$
which is the claim.
