$e^{\left(\pi^{(e^\pi)}\right)}\;$ or $\;\pi^{\left(e^{(\pi^e)}\right)}$. Which one is greater than the other? $\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
$$e^{\bigxl(\pi^{(e^\pi)}\bigxr)}\ge \pi^{\bigxl(e^{(\pi^e)}\bigxr)}$$
 A: We use the following fact in the proof:

Let $c > 0$. Then $\ln(x + c) < \ln(x) + c$ for $x \geq 1$.

For notational convenience, we use the notation $f(x) \rightarrow g(x)$ to denote that $g(x) = \ln f(x)$. We have
$$
e^{\pi^{e^\pi}} \to \pi^{e^\pi} \rightarrow e^\pi\ln \pi \to \pi + \ln\ln \pi
$$
and
$$
\pi^{e^{\color{red}{\pi^e}}} < \pi^{e^{\color{red}{e^\pi}}} \to e^{e^\pi}\ln \pi \to e^\pi + \ln\ln \pi \to \ln(e^\pi + \ln\ln\pi) < \ln(e^\pi) + \ln\ln\pi = \pi + \ln\ln\pi
$$
Therefore,
$$
e^{\pi^{e^\pi}} > \pi^{e^{\pi^e}}
$$
A: Starting from $\pi^e\lt e^{\pi}$, we have, by taking the logarithm twice and doing a trivial bit of algebra,
$$\pi^e\lt e^{\pi}\implies e\ln\pi\lt\pi\implies1+\ln\ln\pi\lt\ln\pi\implies\ln\ln\pi\lt\ln\pi-1$$
We'll use the two ends of the above in the following, which begins by taking a logarithm, then does some trivial algebra, and ends by exponentiating twice:
$$\begin{align}
e\lt\pi&\implies1\lt\ln\pi\\
&\implies e^{\pi}-1\lt(e^{\pi}-1)\ln\pi\\
&\implies e^{\pi}+\ln\pi-1\lt e^{\pi}\ln\pi\\
&\implies\pi^e+\ln\ln\pi\lt e^{\pi}\ln\pi\quad\text{(using }\pi^e\lt e^{\pi}\text{ and }\ln\ln\pi\lt\ln\pi-1)\\
&\implies e^{\pi^e}\ln\pi\lt\pi^{e^{\pi}}\\
&\implies\pi^{e^{\pi^e}}\lt e^{\pi^{e^{\pi}}}
\end{align}$$
A: 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 $b:=\pi$     
