$\text{ Use the function }f\left(x\right)=x^{1/x},x>0,\text{ to determine the bigger of the two }$$\text{ numbers }{e}^{\pi }\text{ and }{\pi }^e\text{. }$

$\begin{array}{l}f\left(x\right)=x^{1/x},(x>0)\\ \Rightarrow f^{\mathrm{\prime }}\left(x\right)=x^{1/x}\frac{(1-1nx)}{x^2}=0\Rightarrow x=e\end{array}$

$\begin{array}{l}\Rightarrow f^{\mathrm{\prime }}\left(x\right)\le 0,\text{ when }x\ge e\\ \Rightarrow f\left(x\right)\text{ decreases when }x\ge e\\ \text{ And }{f}^{\mathrm{\prime }}\left(x\right)>0,\text{ when }0<x<e\\ \Rightarrow f\left(x\right)\text{ increases when }0<x<e\\ \text{ Now }\pi >e\\ \Rightarrow f\left(\pi \right)<f\left(e\right)\left(\text{ as }f\right(x\left)\text{ decreasesin }\right[e,\mathrm{\infty }\left]\right)\\ \Rightarrow {\pi }^{1/\pi }<e^{1/e}\end{array}$

$\text{ Now raise positive power }e\pi \text{ on both sides. }$

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\left({\pi }^{1/\pi }\right)}^{e\pi }<{\left(e^{1e}\right)}^{e\pi }\\ so, \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\pi }^e>e^{\pi }\end{array}$