How can I

compare (without calculator or similar device) the values of $\pi^e$ and $e^\pi$ ?


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 $\pi$.
