Show $\lim_{X \to -k}\:\prod_{j\neq k}\left(X+j \right)=(-1)^{k}k!\left(n-k \right).)$ 
I would like to know how to show that :
  $$\lim_{X \to -k}\:\prod_{j\neq k}\left(X+j \right)=(-1)^{k}k!\left(n-k \right).$$

This is came from solution of exercise that he said :


*

*what is Partial fraction decomposition of :


$$F(X)=\dfrac{n!}{\prod_{k=0}^{n}\left(X+k \right)}$$
indeed,
PFD of F:
$$F(X)=\sum_{k=0}^{n}\dfrac{a_k}{X+k} $$
$$\left(X+k\right)F(X)=\dfrac{n!}{\prod_{j\neq k}\left(X+j \right)}$$
and 
$$\prod_{j\neq k}\left(X+j \right)=(-1)^{k}k!\left(n-k \right)$$ then :
$$a_k=(-1)^{k}{n \choose k} $$
Finaly:
$$F(X)=\sum_{k=0}^{n}(-1)^{k}{n \choose k}\dfrac{1}{X+k} $$ 
 A: I guess you would rather like to prove

$$
\lim_{X \to -k}\:\prod_{j\neq k}\left(X+j \right)=(-1)^{k}k!\left(n-k \right).
$$

If this is the case, one may observe that 
$$
\begin{align}
\lim_{X \to -k}\:\prod_{j\neq k}\left(X+j \right)&=\left.\prod_{j\neq k}\left(X+j \right)\right|_{X=-k}
\\\\&=\prod_{j\neq k}\left(-k+j \right)
\\\\&=\prod_{0\le j \le k,\,j\neq k}\left(-k+j \right)\prod_{k\le j \le n,\,j\neq k}\left(-k+j \right)
\\\\&=(-1)^{k}\prod_{0\le j \le k-1}\left(k-j \right)\prod_{k+1\le j \le n}\left(j-k \right)
\\\\&=(-1)^kk!(n-k)!
\end{align}
$$ as announced.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
\color{#f00}{\mrm{F}\pars{X}} & =
{n! \over \prod_{k = 0}^{n}\pars{X + k}} =
{n! \over \prod_{k = 0}^{n}\pars{k + X}} =
{n! \over \pars{X}_{n + 1}}
\end{align}
where $\ds{\pars{a}_{m}}$ is the Pochhammer Symbol.

Also,
\begin{align}
\prod_{j \not= k}\pars{X + j} & = {n! \over \pars{X + k}\,\mrm{F}\pars{X}} =
{n! \over \pars{X + k}\bracks{n!/\pars{X}_{n + 1}}} =
{\pars{X}_{n + 1} \over X + k} =
{\Gamma\pars{X + n + 1} \over \pars{X + k}\Gamma\pars{X}}
\\[5mm] & =
{\pars{X + n}! \over \pars{X + k}\pars{X - 1}!} =
{\pars{X + n}\ldots X \over X + k} =
\pars{X + n}\ldots\pars{X + k + 1}\pars{X + k - 1}\ldots X
\end{align}

\begin{align}
\lim_{X \to -k}\,\,\,\prod_{j \not= k}\pars{X + j} & =
\pars{-k + n}\pars{-k + n - 1}\ldots 1\bracks{\pars{-1}\ldots\pars{-k}}
\\[5mm] & =
\pars{n - k}!\bracks{\pars{-1}^{k}\,1 \times 2 \ldots k} =
\color{#f00}{\pars{-1}^{k}\,k!\pars{n - k}!}
\end{align}
