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.

I thought it might be instructive to present a proof by induction.  
First, we establish a base case.  For $n=0$, it is straightforward to show that the expression holds.  
Second, we assume that for some $n\geq 0$, we have
$$\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{x+k}=n!\prod_{k=0}^n\frac{1}{x+k}$$
Third, we analyze the sum $\sum_{k=0}^{n+1} \binom{n+1}{k}\frac{(-1)^k}{x+k}$.  Clearly, we can write
$$\begin{align}
\sum_{k=0}^{n+1} \binom{n+1}{k}\frac{(-1)^k}{x+k}&=\frac1x+\sum_{k=1}^{n} \binom{n+1}{k}\frac{(-1)^k}{x+k}+\frac{(-1)^{n+1}}{x+n+1}\\\\
&=\color{green}{\frac1x}+\sum_{k=1}^{n} \left(\color{green}{\binom{n}{k}}+\color{blue}{\binom{n}{k-1}}\right)\frac{(-1)^k}{x+k}+\color{red}{\frac{(-1)^{n+1}}{x+n+1}}\\\\
&=\color{green}{n!\prod_{k=0}^n\frac{1}{x+k}}+\color{blue}{\sum_{k=0}^{n-1} \binom{n}{k}\frac{(-1)^{k+1}}{(x+1)+k}}+\color{red}{\frac{(-1)^{n+1}}{x+n+1}}\\\\
&=\color{green}{n!\prod_{k=0}^n\frac{1}{x+k}}+\color{blue}{-n!\prod_{k=0}^n\frac{1}{x+1+k}-\frac{(-1)^{n+1}}{x+n+1}}+\color{red}{\frac{(-1)^{n+1}}{x+n+1}}\\\\
&=(n+1)!\prod_{k=0}^{n+1}\frac{1}{x+k}
\end{align}$$ 
And we are done!
