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