Simple Isn't it?

1.-Consider the function $f(x) = x^{\frac{1}{x}} = e^{\frac{1}{x}\ln x} \text{ } \forall x > 0,x \in \mathbb{R}$ . Differentiating gives $f '(x) = x^{\frac{1}{x}} \cdot \frac{1}{x^2}(1-lnx) \Rightarrow f '(e) = 0 \text{ and } f '(x) > 0 \text{ }\forall x < e, f '(x) < 0 \text{ } \forall x > e$ , so the function attains its global maximum at x=e.

Thus $e^{\frac{1}{e}} \geq π^{\frac{1}{π}} \Rightarrow (e^{\frac{1}{e}})^{e\pi} = e^{\pi} \geq (π^{\frac{1}{π}})^{e \pi} = \pi^{e}$ and the inequality is clearly strict.

or

2.-If you know Taylor expansion: $e^{x} = 1 + x + \frac{x^2}{2!} +...\Rightarrow$

$e^{x} >1 + x,∀x>0$

Then set $x = \frac{π}{e} -1 >0$

We get $e^{\frac{π}{e}-1} > 1 + \frac{π}{e}-1 ⇔ \frac {e^{\frac {π}{e}}}{e} > \frac {π}{e} ⇔ e^{\frac{π}{e}} >π ⇔ e^π > π^e$