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.

An induction on $n$ will work. Let
$$f(x,n)=\sum_{k=0}^n\frac{(-1)^k}{x+k}\binom{n}k\;,$$
and for the induction step suppose that
$$f(x,n)=\frac{n!}{x(x+1)\ldots(x+n)}\;.$$
Then
$$\begin{align*}
\frac{(n+1)!}{x(x+1)\ldots(x+n+1)}&=\frac{n+1}{x+n+1}f(x,n)\\
&=\frac{n+1}{x+n+1}\sum_{k=0}^n\frac{(-1)^k}{x+k}\binom{n}k\\
&=(n+1)\sum_{k=0}^n(-1)^k\binom{n}k\frac1{(x+k)(x+n+1)}\\
&=(n+1)\sum_{k=0}^n\frac{(-1)^k}{n+1-k}\binom{n}k\left(\frac1{x+k}-\frac1{x+n+1}\right)\\
&=(n+1)\sum_{k=0}^n\frac{(-1)^k}{n+1}\binom{n+1}k\frac1{x+k}\\
&\qquad-\frac{n+1}{x+n+1}\sum_{k=0}^n\frac{(-1)^k}{n+1}\binom{n+1}k\\
&=\sum_{k=0}^n\frac{(-1)^k}{x+k}\binom{n+1}k-\frac1{x+n+1}\color{red}{\sum_{k=0}^n(-1)^k\binom{n+1}k}\\
&=f(x,n+1)-\frac{(-1)^{n+1}}{x+n+1}-\frac1{x+n+1}\color{red}{\left(0-(-1)^{n+1}\right)}\\
&=f(x,n+1)\;,
\end{align*}$$
as desired.
