# $e^{\left(\pi^{(e^\pi)}\right)}\;$ or $\;\pi^{\left(e^{(\pi^e)}\right)}$. Which one is greater than the other?


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

$$e^{\bigxl(\pi^{(e^\pi)}\bigxr)}\ge \pi^{\bigxl(e^{(\pi^e)}\bigxr)}$$

• Roughly speaking, you want big numbers as high in the stack as possible, so the first should be larger. Others have proven this. – Ross Millikan Jul 13 '16 at 21:15
• You should probably change all $\geq$ to $>$, because there is obviously no equality here – Yuriy S Jul 13 '16 at 22:05

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, $$e^{\pi^{e^\pi}} > \pi^{e^{\pi^e}}$$

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}

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 $b:=\pi$

• This is the answer I was looking for! Generalized version, without limiting oneself to $e$ and $\pi$. Thank you :-) – Kusavil Jan 13 '18 at 21:48
• @Kusavil : I am glad that it helps you! :-) – user90369 Jan 14 '18 at 12:16