Contour integral of fraction (involving Gamma function) = Polynomial 
*

*Let $n$ be a natural number ( $0$ included ) and let
$x > 0$ be real.

*Let $\gamma$ be a closed curve which encircles the points $0, 1, 2, \ldots, n$ in positive direction
( anti-clockwise ).
$$
\mbox{Define}\quad\operatorname{P}_{n}\left(x\right) = \frac{1}{2\pi\mathrm{i}}
\oint_{\gamma }\frac{\Gamma\left(t - n\right)}
{\Gamma^{2}\left(t + 1\right)}\,x^{t}\,\mathrm{d}t.
$$
Show that $P_{n}(x)$ is a polynomial of degree $n$.
My work so far: I tried to rewrite $\Gamma(t-n)$ by using the identity $\Gamma(z) = \frac{\Gamma(z+1)}{z}.$ Doing this I obtain that:
$P_{n}(x) = \frac{1}{2\pi i} \int_{\gamma}\frac{x^{t}}{\Gamma(t+1)(t-n)(t-n+1)...(t)}dt$.
From there, I can't see why that expression would be a polynomial. Any hints would be appreciated.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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$\ds{\bbox[5px,#ffd]{}}$

The integrand numerator has poles at $\ds{1, 2, 3, \ldots, n}$: $\color{red}{The\ residue}$ at a pole $\ds{k \in \braces{1,2,3,\ldots,n}}$ is given by:
\begin{align}
&\lim_{t \to k}\bracks{\pars{t - k}
{\Gamma\left(t - n\right) \over
\Gamma^{2}\pars{t + 1}}\,x^{t}}
\\[5mm] = &\
{x^{k} \over
\Gamma^{2}\pars{k + 1}}
\lim_{t \to k}\bracks{\pars{t - k}
\Gamma\pars{t - n}}
\\[5mm] = &\
{x^{k} \over\pars{k!}^{2}}
\lim_{t \to k}\bracks{\pars{t - k}
{\pi \over \Gamma\pars{1 - t + n}
\sin\pars{\pi\bracks{t - n}}}}
\\[5mm] = &\
\pi\,{x^{k} \over\pars{k!}^{2}\pars{n - k}!}
\lim_{t \to k}\
{t - k \over \sin\pars{\pi\bracks{t - n}}}
\\[5mm] = &\
\pi\,{x^{k} \over\pars{k!}^{2}\pars{n - k}!}
\lim_{t \to k}\
{1 \over \cos\pars{\pi\bracks{t - n}}\pi}
\\[5mm] = &\
\pars{-1}^{k - n}\,
{x^{k} \over\pars{k!}^{2}\pars{n - k}!}
\end{align}

Then,
\begin{align}
\operatorname{P}_{n}\pars{x} & \equiv
\bbox[5px,#ffd]{{1 \over 2\pi\ic}
\oint_{\gamma }{\Gamma\pars{t - n} \over \Gamma^{2}\pars{t + 1}}\,x^{t}\,\dd t}
\\[5mm] = &
\bbx{\pars{-1}^{n}\sum_{k = 1}^{n}
{\pars{-1}^{k} \over\pars{k!}^{2}\pars{n - k}!}\,x^{k}}
\\ &
\end{align}
