Limit of an integral related to Gamma function: $ \lim_{k \to \infty}\int_{0}^{k} x^{n}\left(1 - {x \over k}\right)^{k}\,\mathrm{d}x = n! $ I have the following equality:
$$
\lim_{k \to \infty}\int_{0}^{k}
x^{n}\left(1 - {x \over k}\right)^{k}\,\mathrm{d}x = n!
$$
What I think is that after taking the limit inside the integral ( maybe with the help of Fatou's Lemma, I don't know how should I do that yet ), then get
$$
\int_{0}^{k}\lim_{k \to \infty}\left[\,%
x^{n}\left(1 - {x \over k}\right)^{k}\,\right]\mathrm{d}x =
\int_{0}^{\infty}x^{n}\,\mathrm{e}^{-x}\,\mathrm{d}x =
\Gamma\left(n + 1\right) = n!
$$
How can I give a clear proof ?.
 A: By writing
$$
\int_{0}^{k} x^{n}\Big(1-\cfrac{x}{k}\Big)^{k}dx=\int_{0}^{\infty}\mathbf{1}_{[0,k]}(x)\: x^{n}\Big(1-\cfrac{x}{k}\Big)^{k}dx,
$$ observing that
$$
0\le \mathbf{1}_{[0,k]}(x)\: x^{n}(1-\cfrac{x}{k})^{k}\le x^ne^{-x}, \quad x\ge0,
$$ and, for $x\ge 0$,
$$
\lim_{k \rightarrow \infty}\Big(1-\cfrac{x}{k}\Big)^{k}=e^{-x}
$$ one may use the Dominated Convergence Theorem.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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Besides the short $\texttt{@Olivier Oloa}$ answer, another straightforward alternative is based on the Beta Function as follows:

\begin{align}
\lim_{k \to \infty}\int_{0}^{k}
x^{n}\pars{1 - {x \over k}}^{k}\,\dd x &
\,\,\,\stackrel{x/k\ \mapsto\ x}{=}\,\,\,
\lim_{k \to \infty}\bracks{k^{n + 1}\int_{0}^{1}x^{n}\pars{1 - x}^{k}\,\dd x}
\\[5mm] & =
\lim_{k \to \infty}\bracks{k^{n + 1}\,\mrm{B}\pars{n + 1,k + 1}}\qquad
\pars{~\mrm{B}:\ Beta\ Function~}
\\[5mm] & =
\lim_{k \to \infty}\bracks{k^{n + 1}\,
{\Gamma\pars{n + 1}\Gamma\pars{k + 1} \over \Gamma\pars{k + n + 2}}}\qquad
\pars{~\Gamma:\ Gamma\ Function~}
\\[5mm] & =
n!\,\lim_{k \to \infty}\bracks{k^{n + 1}\,
{k! \over \pars{k + n + 1}!}}
\\[5mm] & =
n!\,\lim_{k \to \infty}
{1 \over \bracks{1 + \pars{n + 1}/k}\bracks{1 + n/k}\ldots\pars{1 + 1/k}} =
\bbx{\ds{n!}}
\end{align}
