Comparing $\pi^e$ and $e^\pi$ without calculating them How can I

compare (without calculator or similar device) the values of $\pi^e$ and $e^\pi$ ?

 A: Another visual proof. Recently published in arXiv
A: This is an old chestnut. As a hint, it's easier to consider the more
general problem: for which positive $x$ is $e^x>x^e$?
A: From Proofs without Words.

A: Alternatively, we can compare $e^{1/e}$ and $\pi^{1/\pi}$.
Let $f(x) = x^{1/x}$. Then $f'(x) = x^{1/x} (1 - \log(x))/x^2$. Since $\log(x) > 1$ for $x > e$, we see that $f'(x) < 0$ for $e < x < \pi$. We conclude that $\pi^{1/\pi} < e^{1/e}$, and so $\pi^e < e^\pi$.
The same calculation shows that $f(x)$ reaches its maximum at $e^{1/e}$, and so in general $x^e < e^x$.
A: Yet another line of thought would be this:
$$\begin{align}
e^{\pi} > \pi^e &\iff \pi \ln(e)>e\ln(\pi)\\
&\iff \pi >e\ln(\pi) \\
&\iff \ln(\pi)>\ln(e)+\ln (\ln (\pi)) \\
&\iff \ln(\pi)>1+\ln (\ln (\pi))
\end{align}$$
By concavity of $\ln$, $x-1>\ln(x)$ for all $x\neq 1$. With $x=\ln(\pi)$ we get the inequality wanted.
A: Another proof uses the fact that $\displaystyle \pi \ne e$ and that $e^x > 1 + x$ for $x \ne 0$.
We have $$e^{\pi/e -1} > \pi/e,$$
and so
$$e^{\pi/e} > \pi.$$
Thus,
$$e^{\pi} > \pi^e.$$
Note: This proof is not specific to $\pi$.
A: \begin{align}
&e^\pi>\pi^e \\[5pt]
\iff&\exp(\pi)>\exp(e\log\pi) \\[5pt]
\iff&\pi>e\log\pi \\[5pt]
\iff&\frac{\pi}{\log\pi}>e \\[5pt]
\iff&\frac{\pi}{\log\pi}>\frac{e}{\log e}
\end{align}
The final line is true because the function $\dfrac{x}{\log x}$ is strictly increasing on $[e,\infty) \, ,$ and the result follows.
A: Let $f (x) =$ $x^\frac1x$
Find value of $x$ such that function  gets maximum value 
For this functions for $x=e$ function will get the maximum value 
so $e^\frac1e$ is greater than $\pi^\frac1\pi$
so $e^\pi$ is greater than $\pi ^e$.
A: Elaborating Robin's answer take $f(x) = \log{x} - \frac{x}{e}$.  We have $$f'(x)= \frac{e-x}{xe}$$ Thus $f'(x)>0$ for $0 < x < e$ and $f'(x) <0$ if $x > e$. Consequently, we have $f(x) < f(e)$ if $x \neq e$.
Exercise: Try to prove this using the same methods: $2^{\sqrt{2}} < e$.
A: Hint:
Prove that the function $f(x)=\frac{e^x}{x^e}, x\geq e$ is strictly increasing on the interval $x\in \left [ e,\pi \right ]$. What is $f(e)$ and $f(\pi)$?
A: Denote $n = e^\pi$, $m = \pi^e$ and $s = \log \pi$. Then $\log n = \pi = e^s$ and $\log m = e \log \pi = es$. Then $$\log \frac {n} {m} = \log n - \log m = e (e^{s - 1} - s).$$ By Taylor expansion, we have $$e^{s - 1} = 1 + (s - 1) + \cdots > s.$$ Then $$\log \frac {n} {m} = e (e^{s - 1} - s) > 0.$$ Hence, $n > m$.
A: Just adding a recent one. Another visual proof. I am doubtful if it has open access.
 https://link.springer.com/article/10.1007/s00283-019-09964-x 
A: Not that this question needs another answer, but here is a proof of $e^\pi > \pi^e$ using the Mean Value Theorem applied to $\ln x$ on the interval $(e,\pi)$, along with the assumptions that $e<\pi$, $\ln$ is increasing, and $\frac{d}{dx} \ln x = \frac{1}{x}$.
By the MVT, there exists $c \in (e,\pi)$ such that
$$ \frac{\ln \pi - \ln e}{\pi - e} = \frac{1}{c} $$
We can increase the right hand side by replacing $c$ with the smaller number $e$, and so we have
$$ \frac{\ln \pi - \ln e}{\pi - e} < \frac{1}{e} $$
and thus
$$ {\ln \pi - \ln e} < \frac{1}{e}(\pi-e) $$
or
$$ {\ln \pi - 1} < \frac{\pi}{e}-1 $$
which gives
$$ e \ln \pi < \pi \ln e$$
and therefore
$$ e^\pi < \pi^e$$
A: Let
$$f(x) = e^x$$
$$G(x) = x^e$$
We can simply show that
$$f(e)=G(e)$$
$$f'(e)=G'(e)$$
For $x > e$ the $f(x)$ will grow faster than $G(x)$
Then
$$e^{\pi} > \pi^{e}$$
