Comparing $e^{\pi}$ and $\pi^e$.

Since $e^x > 1+x$ for positive $x$ (proof: $\frac{d}{dx}e^x \ge \frac{d}{dx}(1-x)$ and $e^x = 1-x$ at $x=0$), if we set $x=\pi/e-1 > 0$,

Generally, for real $x$, we have $e^x \ge 1+x$ with the equality holds only at $x=0$. Therefore,

and the equality holds only at $x/e-1=0$ or $x=e$.

Alternative way to prove $e^{\pi} > \pi^e$ is to compare $e^{1/e}$ and $\pi^{1/\pi}$. Consider