Alternating sum over binomial coefficients I am trying to follow a proof in a physics paper, but got stuck with the identity
$$\sum_{i=0}^n(-1)^i\binom{k}{n-i}\frac{(m+i)!}{i!} = m!\binom{k-m-1}{n}.$$
I would be very grateful if you could shed light on this mystery. 
 A: Here we  have   Chu-Vandermonde's Identity in disguise.

Dividing the left hand side by $m!$ we obtain
  \begin{align*}
\color{blue}{\sum_{i=0}^n}&\color{blue}{(-1)^i\binom{k}{n-i}\frac{(m+i)!}{i!m!}}\\
&=\sum_{i=0}^n(-1)^i\binom{k}{n-i}\binom{m+i}{i}\\
&=\sum_{i=0}^n\binom{k}{n-i}\binom{-m-1}{i}\tag{1}\\
&\,\,\color{blue}{=\binom{k-m-1}{n}}\tag{2}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

*In (2)  we apply the Chu-Vandermonde identity.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{i = 0}^{n}\pars{-1}^{i}{k \choose n - i}{\pars{m + i}! \over i!} = m!{k - m - 1 \choose n}:\ {\LARGE ?}}$.

\begin{align}
&\bbox[10px,#ffd]{\sum_{i = 0}^{n}\pars{-1}^{i}{k \choose n - i}
{\pars{m + i}! \over i!}} =
\sum_{i = 0}^{\infty}\pars{-1}^{i}\,\,\,
\overbrace{\bracks{z^{n - i}}\pars{1 + z}^{k}}^{\ds{k \choose n - i}}\,\,\,
\overbrace{m!{m + i \choose i}}^{\ds{\pars{m + i}! \over i!}}
\\[5mm] = &\
m!\bracks{z^{n}}\pars{1 + z}^{k}\sum_{i = 0}^{\infty}{m + i \choose i}
\pars{-z}^{i}
\\[5mm] = &\
m!\bracks{z^{n}}\pars{1 + z}^{k}\sum_{i = 0}^{\infty}
\overbrace{{-m - i + i - 1 \choose i}\pars{-1}^{i}}^{\ds{m + i \choose i}}
\,\,\,\pars{-z}^{i}
\\[5mm] = &\
m!\bracks{z^{n}}\pars{1 + z}^{k}\sum_{i = 0}^{\infty}
{-m - 1 \choose i}z^{i} =
m!\bracks{z^{n}}\pars{1 + z}^{k}\pars{1 + z}^{-m - 1}
\\[5mm] = &\
m!\bracks{z^{n}}\pars{1 + z}^{k - m - 1} =
\bbx{m!{k - m - 1 \choose n}}
\end{align}
