TITLE: $e^{\left(\pi^{(e^\pi)}\right)}\;$ or $\;\pi^{\left(e^{(\pi^e)}\right)}$. Which one is greater than the other?
QUESTION [16 upvotes]: $\newcommand{\bigxl}[1]{\mathopen{\displaystyle#1}}
\newcommand{\bigxr}[1]{\mathclose{\displaystyle#1}}
$
$$\large e^{\bigxl(\pi^{(e^\pi)}\bigxr)}\quad\text{or}\quad\pi^{\bigxl(e^{(\pi^e)}\bigxr)}$$
Which one is greater?

Effort. I know that $$e^\pi\ge \pi^e$$
Then $$\pi^{(e^\pi)}\ge e^{(\pi^e)}$$
But I can't say
$$e^{\bigxl(\pi^{(e^\pi)}\bigxr)}\le \pi^{\bigxl(e^{(\pi^e)}\bigxr)}$$
or
$$e^{\bigxl(\pi^{(e^\pi)}\bigxr)}\ge \pi^{\bigxl(e^{(\pi^e)}\bigxr)}$$

REPLY [4 votes]: The function $x^{\frac{1}{x}}$ is strictly decreasing for $x>e$, the maximum is $e^{\frac{1}{e}}$. 
Therefore: 
$e\le a<b$ => $a^{\frac{1}{a}}>b^{\frac{1}{b}}$ => $a^b>b^a$ => $a^b-1>b^a-1$ => $(a^b-1)\ln b>(b^a-1)\ln a$ 
=> $b^{a^b-1}>a^{b^a-1}$ 
=> $b^{a^b-1}\frac{\ln a}{a}>a^{b^a-1}\frac{\ln b}{b}$ => $a^{b^{a^b}}>b^{a^{b^a}}$ 

Here: $a:=e$ and $b:=\pi$