Which is larger, e^π or π^e?

1. $h'(x)=\dfrac{1-\ln x}{x^2}$, positive on $(0,e)$ and negative on $(e,+\infty)$. So $h$ increases up to $x=e$ and then decreases, with maximum $h(e)=\frac1e$.

2. For all $x>0$, $\ln\!\left(x^{1/x}\right)=h(x)\le\frac1e=\ln\!\left(e^{1/e}\right)$. Since $\exp$ is increasing, $x^{1/x}\le e^{1/e}$, with equality only at $x=e$.

3. Since $\pi>e$ and $h$ is strictly decreasing on $[e,+\infty)$, $h(\pi)<h(e)$, that is
$$\frac{\ln\pi}\pi<\frac1e\iff e\ln\pi<\pi\iff\ln\left(\pi^e\right)<\ln\left(e^\pi\right).$$
Therefore $\pi^e<e^\pi$.

Numerically. $\pi^e\approx22.46$ while $e^\pi\approx23.14$: they are close, but the order is clear.

Remark. The same reasoning shows that $a^b>b^a$ whenever $e\le a<b$. For example $3^4=81>4^3=64$.