So my text book on probability theory gives an example of a distribution with random parameters, which after a few paragraphs ends up with this integral:
$$ \begin{split} \int_0^{\infty}\frac{x^k}{k!}e^{-2x}dx &= \frac{1}{2^{k+1}} \int_0^{\infty}\frac{1}{\Gamma(k+1)}2^{k+1}x^{k+1-1}e^{-2x}dx\\ &= \frac{1}{2^{k+1}} \cdot 1 \end{split} $$
where $\Gamma$ is the Gamma function.
My question is this: How in the world did the author calculate this integral?
For those of you who are curious, that integral is the result of calculating $P(X=k|M=x)$ with $M \in \text{Exp}(1)$ and $X|M=m \in \text{Po}(m)$. The example can be found in An intermediate course in Probability, 2nd Ed. (Gut, 2009, pp 39).
Just doing the substitution $u=2x$ and using $$\Gamma (t)=\int_0^\infty x^{t-1}e^{-x}\,\mathrm d x.$$