Prove that $\sum^n_{k=0}\frac{(-1)^k}{k+x}\binom{n}{k}=\frac{n!}{x(x+1)\cdots(x+n)}$. Given the following formula 
$$
\sum^n_{k=0}\frac{(-1)^k}{k+x}\binom{n}{k}\,.
$$
How can I show that this is equal to 
$$
\frac{n!}{x(x+1)\cdots(x+n)}\,?
$$
 A: Consider the (unique) polynomial $p(x)\in\mathbb{Q}[x]$ of degree at most $n$ such that $p(-k)=1$ for all $k=0,1,2,\ldots,n$.  Clearly, $p(x)$ is the constant polynomial $1$.  
However, using Lagrange interpolation, we have
$$p(x)=\sum_{k=0}^n\,p(-k)\,\frac{\prod\limits_{j\in[n]\setminus\{k\}}\,(x+j)}{\prod\limits_{j\in[n]\setminus\{k\}}\,(-k+j)}\,,$$
where $[n]:=\{0,1,2,\ldots,n\}$.  This means
$$1=\sum_{k=0}^n\,\frac{\prod\limits_{j\in[n]\setminus\{k\}}\,(x+j)}{\prod\limits_{j\in[n]\setminus\{k\}}\,(-k+j)}=\sum_{k=0}^n\,(-1)^k\,\frac{\prod\limits_{j\in[n]\setminus\{k\}}\,(x+j)}{k!\,(n-k)!}\,.$$
Multiplying both sides by $\dfrac{n!}{\prod\limits_{j\in[n]}\,(x+j)}$ yields
$$\frac{n!}{\prod\limits_{j=0}^n\,(x+j)}=\sum_{k=0}^n\,(-1)^k\,\left(\frac{n!}{k!\,(n-k)!}\right)\,\frac{1}{x+k}=\sum_{k=0}^n\,\binom{n}{k}\,\frac{(-1)^k}{x+k}\,.$$
A: Induction step:
$$\begin{align}
\sum_{k=0}^{n+1}&\frac{(-1)^k}{x+k}\binom{n+1}k=\frac1x+\frac{(-1)^{n+1}}{x+n+1}+\sum_{k=1}^{n}\frac{(-1)^k}{x+k}\left[\binom nk+\binom n{k-1}\right]
\\&=\frac{n!}{x(x+1)\cdots(x+n)}+\frac{(-1)^{n+1}}{x+n+1}+\sum_{k=1}^{n}\frac{(-1)^k}{x+k}\binom{n}{k-1}
\\&=\frac{n!}{x(x+1)\cdots(x+n)}+\frac{(-1)^{n+1}}{x+n+1}-\sum_{k=0}^{n-1}\frac{(-1)^k}{(x+1)+k}\binom{n}{k}
\\&=\frac{n!}{x(x+1)\cdots(x+n)}+\frac{(-1)^{n+1}}{x+n+1}-\frac{n!}{(x+1)(x+2)\cdots(x+n+1)}+\frac{(-1)^n}{x+n+1}
\\&=\frac{n!(x+n+1)-n!x}{x(x+1)\cdots(x+n+1)}=\frac{(n+1)!}{x(x+1)\cdots(x+n+1)}
\end{align}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{\Re\pars{x} > 0}$:

\begin{align}
&\bbox[10px,#ffd]{\sum_{k = 0}^{n}{\pars{-1}^{k} \over k + x}{n \choose k}} =
\sum_{k = 0}^{n}\pars{-1}^{k}
\pars{\int_{0}^{1}t^{k + x - 1}\,\dd t}{n \choose k}
\\[5mm] = &\
\int_{0}^{1}t^{x - 1}\sum_{k = 0}^{n}
{n \choose k}\pars{-t}^{k}\,\dd t
\\[5mm] = &\
\int_{0}^{1}t^{x - 1}\,\pars{1 - t}^{n}\,\dd t =
\mrm{B}\pars{x,n + 1}\ \pars{~\mrm{B}:\ Beta\ Function~}
\\[5mm] = &\
{\Gamma\pars{x}\Gamma\pars{n + 1} \over \Gamma\pars{x + n + 1}}
\phantom{= \mrm{B}\pars{x,n + 1}\,\,\,\,\,\,\,\,\,\,\,\,}
\pars{~\Gamma:\ Gamma\ Function~}
\\[5mm] = &\
{n! \over \Gamma\pars{x + n + 1}/\Gamma\pars{x}} =
{n! \over x^{\overline{n +1}}} =
\bbx{n! \over x\pars{x + 1}\cdots\pars{x + n}}
\end{align}
