How can I calculate, without calculator or similar device,
the values of $\pi^e$ and $e^\pi$
in order to compare them?
How can I calculate, without calculator or similar device,
the values of $\pi^e$ and $e^\pi$
in order to compare them?
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 $\pi$.
This is an old chestnut. As a hint, it's easier to consider the more general problem: for which positive $x$ is $e^x>x^e$?
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 $x^e < e^x$.
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 $\pi ^e$.
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: $2^{\sqrt{2}} < e$.
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 $f(\pi)$?
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, $n > m$.
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.
$$e^\pi>\pi^e$$ because if we subtract $e$ from both exponents we get $$e^{\pi-e}>1$$ which is true because $e$ is is greater than $1$ so when it 8s raised to that power it is equal to $$\frac{e^\pi}{e^e}$$ and that has to be greater than 1 because the top is greater than the bottom.