How to prove $\lfloor{\pi}\rfloor+\lfloor{\pi^e}\rfloor=\lfloor{e}\rfloor+\lfloor{e^\pi}\rfloor$? 
How to prove $\lfloor{\pi}\rfloor+\lfloor{\pi^e}\rfloor=\lfloor{e}\rfloor+\lfloor{e^\pi}\rfloor$?(without using the calculator)

$e$ is Euler's number.
I tried to solve this question. But I could only get to the following inequality:
$e^\pi >\pi^e$. I got this inequality by deriving from the function $f(x)=e^x-x^e$. so $f'(x) \geq 0$ for $x > e$. This means that $f$ is an increasing function, and $f(e) = 0$.Thus, $f(\pi) > 0$, so $e^{\pi} - \pi^{e} > 0$. This implies that $e^{\pi}$ is greater than $\pi^{e}$.
 A: This is not a full answer, however I do explain one approach to establishing the value of the RHS, and at the end I will give an observation that hopefully someone can build off of to prove the LHS.
For a start, I think it is pretty well established that $\lfloor e\rfloor=2$ and $\lfloor\pi\rfloor=3$. These follow immediately from continued-fraction expansions and other well-known formulas.
To find a lower-bound on $e^\pi$, we can use the exponential taylor series combined with a well-known approximation $\pi>\frac{339}{108}$. So, we find that
$$e^\pi>\sum_{n=0}^\infty\frac{1}{n!}\left(\frac{339}{108}\right)^n.$$
Although tedious, truncating after $n=9$  terms, we thus find
$$e^\pi>\frac{849221956019853023669}{36854077075834798080},$$
which is not hard to show is greater than $23$.
Furthermore, we can establish an upper bound by noting that $\pi<\frac{256}{81}$, and thus
$$\sum_{n=0}^\infty\frac{1}{n!}\left(\frac{256}{81}\right)^n>e^\pi,$$
which can be shown to be greater than 23 (by truncating after $n=7$), but the remaining terms can never account for a difference that brings the full sum past $24$ (which is difficult to explain in brevity, but very doable).
Thus,
$$23<e^\pi<24\implies\lfloor e^\pi\rfloor=23\implies\lfloor e\rfloor+\lfloor e^\pi \rfloor=25$$
Establishing bounds for $\pi^e$ is more difficult however. As pointed out, it may be easier to work with $\ln\pi$. One idea is to make use of the fact that
$$\zeta'(0)=-\ln\sqrt{2\pi},$$
$$\implies \ln\pi = -\ln 2-2\zeta'(0).$$
From there, one could use the many series and approximations for $\ln(2)$ and $\zeta'$, respectively.
