Which is larger \(\pi^e\) or \(e^{\pi}\)? Hint: consider \(\log x/x\).

Suppose \(\alpha>\beta\). The key observation is if we take the natural logarithm of both sides, then \(\log \alpha > \log \beta\) would imply \(\alpha>\beta\) since the logarithm is montonically increasing (and concave).

Hence, let \(y=\frac{\log x}{x}\). Consider the first derivative: \begin{equation} \frac{dy}{dx}=\frac{1}{x^2}-\frac{\log}{x^2}=\frac{1-\log x}{x^2}. \end{equation}

Note that \(y'(e)=\frac{1-\log e}{e^2}=0\) and \(y'(\pi)=\frac{1-\log \pi}{\pi^2}<0\), as \(\pi > e\). Indeed, \(y'(x)<0\) on \([e,\pi]\). Notice that \(y(x)\) has a maximum at \(x=e\) since \(y'(e)=0\). Thus, \begin{equation} \frac{\log e}{e} > \frac{\log \pi}{\pi}. \end{equation}

It follows that \(\pi \log e>e\log \pi\) or \(e^\pi>\pi^e\).

One can also use a Taylor series for \(e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}\), substituting \(x=\frac{\pi}{e}-1\) to see this, albeit less intutivie.