Let $f(x)=e^x-x^e$, and hence $f\text{'}(x)=e^x-e{x}^{e-1}=e(e^{x-1}-x^{e-1})$

$f\text{'}(x)=0$ when $x=1$ or $e$.

$f(0)=1$ and $f\text{'}(0)=1$

As $x$ tends to infinity, $f(x)$ tends to infinity.

$f(1)=e-1$ and $f(e)=0$.

So, assuming that the only points $x>0$ where $f\text{'}(x)=0$ are $x=1$ and $x=e$, the graph of $f$ looks like: it starts at $f=1$ when $x=0$, increases up to $f=e-1$ when $x=1$, decreases down to $f=0$ when $x=e$, and then increases up to infinity as $x$ increases beyond that.

So, $f(x)\ge 0$ for all $x$, and $f(x)=0$ only if $x=e$. In other words $f(\pi )>0$, so that $e^{\pi }-{\pi }^e>0$, so $e^{\pi }>{\pi }^e$. The only thing you need to check is that the only positive real zeroes of $f\text{'}(x)$ are $x=1$ and $x=e$. In fact, $e^{\pi }\approx 23.14$ and ${\pi }^e\approx 22.46$.