Prove the identity $ \sum\limits_{k=0}^{n-1} (-1)^{k} \binom{n}{k} (n-k) \frac{(a+k-2)!}{(a+k-n)!}=0.$ Prove the identity 
$$
\sum_{k=0}^{n-1} (-1)^{k}  \binom{n}{k} (n-k)  \frac{(a+k-2)!}{(a+k-n)!}=0
$$
I want to reduce
$$
\frac{(a+k-2)!}{(a+k-n)!}=(2-n)\binom{a+k-n}{a+k-2}^{-1}=(2-n) (a+k-2) \int_0^1 z^{a+k-2}(1-z)^{2-n} dz,
$$ 
and then try integrate it 
but the binomial coefficient 
$$
\binom{a+k-n}{a+k-2}=\frac{(a+k-n)!}{(2-n)! (a+k-2)!},
$$
is undefined for $n>2$ however the sum seems  correct for any $n.$
Any ideas?
 A: 
We obtain for integral $n>0$ and $a\in\mathbb{C}\setminus\{n-1,n-2,n-3,\ldots\}$
  \begin{align*}
\color{blue}{\sum_{k=0}^{n-1}}&\color{blue}{(-1)^k\binom{n}{k}(n-k)\frac{(a+k-2)!}{(a+k-n)!}}\\
&=\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}k\frac{(a+n-k-2)!}{(a-k)!}\tag{1}\\
&=(-1)^nn(n-2)!\sum_{k=1}^n(-1)^k\binom{n-1}{k-1}\binom{a+n-k-2}{a-k}\tag{2}\\
&=(-1)^{n+1}n(n-2)!\sum_{k=0}^{n-1}\binom{n-1}{k}\binom{a+n-k-3}{a-k-1}\tag{3}\\
&=(-1)^{n+1}n(n-2)!\sum_{k=0}^{n-1}\binom{n-1}{k}\binom{-n+1}{a-k+1}\tag{4}\\
&=(-1)^{n+1}n(n-2)!\sum_{k=0}^{n-1}\binom{n-1}{k}[z^{a-k+1}](1+z)^{-n+1}\tag{5}\\
&=(-1)^{n+1}n(n-2)![z^{a+1}](1+z)^{-n+1}\sum_{k=0}^{n-1}\binom{n-1}{k}z^k\tag{6}\\
&=(-1)^{n+1}n(n-2)![z^{a+1}](1+z)^{-n+1}(1+z)^{n-1}\tag{7}\\
&=(-1)^{n+1}n(n-2)![z^{a+1}]1\\
&\,\,\color{blue}{=0}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we set the upper limit of the sum to $n$ (i.e. adding zero) and change the order of summation by $k\to n-k$.

*In (2) we apply the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$ and use binomial coefficients.

*In (3) we shift the index to start with $k=0$.

*In (4) we apply the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^{q}$.

*In (5) we use the coefficient of operator $[z^p]$ to denote the coefficient of a series. This way we can write 
$[z^p](1+z)^q=\binom{q}{p}$.

*In (6) we use the linearity of the coefficient of operator and apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

*In (7) we apply the binomial theorem.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&{1 \over \pars{n - 2}!}\bbox[10px,#ffd]{\ds{\sum_{k = 0}^{n - 1}\pars{-1}^{k}{n \choose k}\pars{n - k}{\pars{a + k - 2}! \over \pars{a + k - n}!}}} =
\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}\pars{n - k}
{a + k - 2 \choose a + k - n}
\\[5mm] = &\
\sum_{k = 0}^{n}\pars{-1}^{n - k}{n \choose k}k
{a + n - k - 2 \choose a - k} =
\sum_{k = 0}^{n}\pars{-1}^{n - k}{n \choose k}
{-n + 1\choose a - k}\pars{-1}^{a - k}\, k
\\[5mm] = &\
\pars{-1}^{n + a}\sum_{k = 0}^{n}{n \choose k}k
\bracks{z^{a - k}}\pars{1 + z}^{-n + 1} =
\pars{-1}^{n + a}\bracks{z^{a}}\pars{1 + z}^{-n + 1}
\sum_{k = 0}^{n}{n \choose k}k\,z^{k}
\\[5mm] = &\
\pars{-1}^{n + a}\bracks{z^{a}}\pars{1 + z}^{-n + 1}\,
z\partiald{}{z}\sum_{k = 0}^{n}{n \choose k}z^{k} =
\pars{-1}^{n + a}\bracks{z^{a - 1}}\pars{1 + z}^{-n + 1}\,
\partiald{\pars{1 + z}^{n}}{z}
\\[5mm] = &\
\pars{-1}^{n + a}\bracks{z^{a - 1}}\pars{1 + z}^{-n + 1}\,
\bracks{n\pars{1 + z}^{n - 1}} =
\pars{-1}^{n + a}\, n\bracks{z^{a - 1}}1 =
\pars{-1}^{n + 1}\, n\bracks{a = 1}
\end{align}

$$
\bbx{\bbox[10px,#ffd]{\ds{\sum_{k = 0}^{n - 1}\pars{-1}^{k}{n \choose k}\pars{n - k}{\pars{a + k - 2}! \over \pars{a + k - n}!}}} =
\bracks{a = 1}\pars{-1}^{n + 1}\,n\pars{n - 2}!}
$$
