Comparing $\pi^e$ and $e^\pi$ without calculating them
Another proof uses the fact that $\displaystyle \pi \ne e$ and that $e^x > 1 + x$ for $x \ne 0$.
We have $$e^{\pi/e -1} > \pi/e,$$
and so
$$e^{\pi/e} > \pi.$$
Thus,
$$e^{\pi} > \pi^e.$$
Note: This proof is not specific to $\pi$.
This is an old chestnut. As a hint, it's easier to consider the more general problem: for which positive $x$ is $e^x>x^e$?