Help me prove $\pi^{e}<23I am a high school student.
Help me prove $\pi^{e}<23<e^{\pi}$.
First, I tried it by proving $\frac{\log \pi}{\pi}<\frac{\log 23}{\pi e}<\frac{\log e}{e}$.
However, I was not able to prove the magnitude relationship between $23$ and the others.
Second, I attemted to approach by quadratic curve at $x=3$.
$e^{x}=e^{3}+e^{3}(x-3)+\frac{1}{2}e^{3}(x-3)^2$.
And the third derivative of $e^{x}$ is positive.So,
$e^{\pi}>\frac{1}{2}e^{3}(\pi^{2}-4\pi +5)> 
\frac{1}{2}*2.717^{3}(3.14159^{2}-4*3.1416+5)>23$
This is how I can prove the supremum of the inequality, but I can't the other.
I ended up stuck here.
Moveover, I'm going so far as to believe there is a more beautiful, plain, and concise proof for this...
 A: For the first part we use $\pi<{22\over7}$. Furthermore the theorem on alternating series gives
$${1\over e}>1-1+{1\over2}-{1\over6}+{1\over24}-{1\over120}={44\over120}={11\over30}\ .$$
It follows that $$\pi^e<\left({22\over7}\right)^{30/11}\ .$$
It is therefore sufficient to prove that 
$$22^{30}<7^{30}\cdot 23^{11}\ .$$
Here the LHS computes to
$$18\,736\,153\,019\,903\,829\,443\,036\,278\,993\,864\,332\,673\,024\ ,$$
and the RHS to
$$21\,475\,703\,365\,914\,111\,444\,329\,770\,286\,387\,088\,568\,823\ .$$
A: Both $\pi\approx\frac{22}{7}$ and $e\approx\frac{19}{7}$ can be proved through Beuker-like integrals:
$$ \underbrace{\int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}\,dx}_{\in\left(0,\frac{1}{256}\right)}=\frac{22}{7}-\pi,\qquad \underbrace{\int_{0}^{1}x^2(1-x)^2 e^{-x}\,dx}_{\in\left(0,\frac{1}{16}\right)}=14-\frac{38}{e} $$
and $\pi^e < e^\pi$ is trivial from the increasing nature of $\frac{x}{\log x}$ on $[e,+\infty)$.
$\pi^e<23$ is actually a loose inequality, since $\left(\frac{22}{7}\right)^{19}\ll 23^7$.
$23<e^\pi$ is a bit tighter, since $23^7\approx 3.4\cdot 10^9$ while $\left(\frac{19}{7}\right)^{22}\approx 3.47\cdot 10^9$.
Still, $\pi^e<23<e^\pi$ is pretty straightforward to prove by approximating both $e$ and $\pi$ in sevenths.
