Which one is $\color{#CEBB00}{larger}$

After noting first that $\large e^{x} \gt x + 1$ for $x \gt 0$ we see that

$\large e^{\frac{\pi}{e} - 1} \gt \left(\dfrac{\pi}{e} - 1 \right) + 1 = \dfrac{\pi}{e} \Longrightarrow e^{\frac{\pi}{e}} \gt \pi \Longrightarrow \boxed{e^{\pi} \gt \pi^{e}}$ .

$\text{e}^{\pi}$ $\approx{23.1406926328}$

$\pi^{e}$ $\approx{22.4591577184}$

$\text{ 23.1406926328} {>} { 22.4591577184}$

Thus the answer is $\text{e}^{\pi}$ $\boxed{>}$ $\pi^{e}$

I'm using stationary knowledge. $\pi^{e} = 22.4591577183$ $e^{\pi} = 23.1406926327$

So, clearly, $e^{\pi} \boxed {>} \pi^{e}$ .

[N.B. If any Bangladeshi problem solver is seeing this, he/she should know that I took this data from "Neurone Anuranan". It was written by respectable Muhammad Zafar Iqbal Sir and Mohammad Kaykobad Sir.]