The boundedness of $x^k e^{-|x|}$ What is a simple way of showing that: for every $k$, there exists $M$ such that
$$\sup_{x \in \mathbb{R}} x^ke^{-|x|}< M?$$
 A: Although I'm a total amateur, I have somewhat of an idea:
Differentiating the function $$f(x) = x^ke^{-|x|}$$ gives
$$f'(x) = kx^{k-1}e^{-|x|} - x^ke^{-|x|}\frac{|x|}{x}$$
Which yields relative mins/maxes only at the points $x = 0,k,-k$
The fact that it has no asymptotes and its limit goes to $0$ take care of any other possible behavior.
Hope this helped. 
A: It seems the following. 
If $k$ is not integer then we have a problem to define $x^k$ for a negative $x$.
If $k<0$ then we have $\lim_{x\to +0} x^ke^{|x|}=\infty$.
If $k$ is integer and $k\ge 0$ then  for each $x\in\mathbb R$ we have $$x^ke^{-|x|}\le |x|^ke^{-|x|}=k!e^{-|x|}\left(\frac{|x|^k}{k!}\right)\le k!e^{-|x|}e^{|x|}=k!.$$
Remark. If we compare this upper bound with the timur's remark to user2303321's answer then we also easily obtain the inequality $k!\ge k^ke^{-k}$, which is a relatively good lower bound for $k!$, as shows Stirling's formula.   
A: The simplest I have seen is this:
If $x \ge 0$,
for any $n$,
$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}
> \frac{x^{n}}{n!}
$
so
$x^n e^{-x} < n!
$.
