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 $$x=\pi/e-1 > 0$$,

\begin{aligned} e^{\pi/e-1} &> 1+\pi/e - 1 = \pi/e \\ e^{\pi/e} &> \pi \\ e^{\pi} &> \pi^e \end{aligned}

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

\begin{aligned} e^{x/e-1} &\ge x/e \\ e^{x/e} &\ge x \\ e^x &\ge x^e \end{aligned}

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

\begin{aligned} f(x) &= x^{1/x} = e^{(1/x)\log(x)} \\ f'(x) &= e^{(1/x)\log(x)}\cdot(\frac{1}{x}\frac{1}{x}-\frac{1}{x^2}\log(x)) \\ &= x^{1/x}\frac{1}{x^2}(1-\log x) \\ \therefore,\quad f'(x) &>0 \textrm{ for } x\in(0,e) \\ f'(x) &=0 \textrm{ for } x=e \\ f'(x) &<0 \textrm{ for } x\in(e,\infty) \\ \textrm{since }\pi>e\textrm{, we have }f(e) &>f(\pi) \\ e^{1/e} &> \pi^{1/\pi} \\ e^{\pi} &> \pi^e. \end{aligned}