[Mastermander]

Since
$$e^x=1+x+\frac{x^2}{2!}+...$$
we get,
$$e^x-1=x+\frac{x^2}{2!}+...$$
when $x>0$ we get $e^x-1>x$ ,
So put $x=\frac{\pi}{e}-1$
as a result $$e^\pi>\pi^e$$