Showing that $x^me^{-ax}$ is bounded Let $a>0$, $m\geq 0$ and let $f:[0,\infty)\rightarrow\mathbb{R}$ be defined as $f(x)=x^me^{-ax}$. It should be true that this function is bounded, because near $0$ both terms are bounded, and far from $0$ the term $e^{-ax}$ decreases more rapidly than the term $x^m$ increases. What is a nice way to show this rigorously?

We can bound instead $g(x) = f(\frac{m}{a}x) \left (\frac{a}{m} \right )^m$
Then we get
\begin{align}
g(x) = \frac{x^m}{ e^m} &\leq \\
\frac{x^m}{\left (e^x \right )^m} &\leq\\
x^m \frac{1}{(\sum _{k\geq 0} \frac{x^k}{k!} )^m }&\leq\\
x^m \frac{1}{(x+1)^m}&\leq 1
\end{align}
