Integral form of $\Gamma (x)$ While trying to represent the poles and analytic continuation of $\Gamma (x)$ , the author uses the following equality:

$$\int\nolimits_{0}^{1}t^{x-1}e^{-t}dt=\sum_{n=0}^{+\infty}\frac{{(-1)}^{n}}{(n+x)n!}.$$
Any help in figuring out how?
In order to expand my comment: put for $0\leq t\leq 1$: $\displaystyle S_n(t)=\sum_{k=0}^n\frac{(-t)^k}{k!}$. We have 
$$\sup_{0\leq t\leq 1}|\exp(-t)-S_n(t)| = \sup_{0\leq t\leq 1}\sum_{k=n+1}^{+\infty}\frac{(-t)^k}{k!}\leq \sum_{k=n+1}^{+\infty}\frac 1{k!},$$
and since the series $\displaystyle\sum_{k=0}^{+\infty}\frac 1{k!}$ is convergent, the sequence $\{S_n\}$ converges uniformly to $t\mapsto \exp (-t)$ on $\left[0,1\right]$. Therefore, we can switch the series and the integral for $x\geq 1$:

$\begin{align*}
\int_0^1t^{x-1}e^{-t}dt&amp;=\int_0^1t^{x-1}\left(\sum_{k=0}^{+\infty}\frac{(-1)^kt^k}{k!}\right)dt\\
&amp;=\sum_{k=0}^{+\infty}\frac{(-1)^k}{k!}\int_0^1 t^{x+k-1}dt\\
&amp;=\sum_{k=0}^{+\infty}\frac{(-1)^k}{k!(x+k)}.
\end{align*}$

For $0&lt;x&lt;1$, we can use the dominated convergence theorem since $|t^{x-1}S_n(t)|\leq e\cdot t^{x-1}$ which is integrable on $\left[0,1\right]$ (we can use a more simple argument in the case $x\geq 1$).
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