Is this inequality provable? $e^{\left(\pi^{e^{\pi^{.^{.^{.^{e^\pi}}}}}}\right)}\ge \pi^{\left(e^{\pi^{e^{.^{.^{.^{\pi^e}}}}}}\right)}$

Define the sequences $E_n$ and $P_n$ by

$$E_{n+1}=e^{\pi^{E_n}}\qquad\text{and}\qquad P_{n+1}=\pi^{e{^{P_n}}}$$

with $E_1=e^{\pi}$ and $P_1=\pi^e$. We know that $E_1\gt P_1$ from elementary calculus considerations (or simply by direct numerical calculations). We wish to show that $E_n\gt P_n$ for all $n$. Doing so takes care of towers of even length. But it also takes care of odd-length towers as well, since $E_n\gt P_n$ and $\pi\gt e$ together imply $\pi^{E_n}\gt e^{P_n}$.

Note that

$$E_{n+1}\gt P_{n+1}\iff\ln\ln E_{n+1}\gt\ln\ln P_{n+1}\iff E_n\ln\pi\gt P_n+\ln\ln\pi$$

Thus to finish off a proof by induction, it suffices to show that

$$E_n(\ln\pi-1)\gt\ln\ln\pi$$

for all $n$. But $E_1\gt1$ kicks off a mini-induction $E_{n+1}=e^{\pi^{E_n}}\ge e^{\pi^{E_1}}\gt e^{\pi^1}= E_1$ for all $n$, so it's enough to show

$$e^{\pi}(\ln\pi-1)\gt\ln\ln\pi$$

There might be some slick analytic way to establish this inequality without any messy calculation, but for the moment at least, let's just note that the two sides aren't even close: $e^\pi(\ln\pi-1)\approx3.3491498$, while $\ln\ln\pi\approx0.1351687$.

Remark: I skipped over the proof that $E_1\gt P_1$ in part because the OP, in the linked-to related question, accepts it as known, so the primary interest here seems to be the higher tetrations. I and others gave purely analytic (non-computational) answers there to the inequality I'm expressing here as $E_2\gt P_2$. Those answers' approaches might might generalize to all subscripts $n$, but offhand I don't see how. That's why I took a partly computational approach here.

We can use the function $y = x^{{1\over x}}$ to prove $e^{\pi} \gt \pi ^e$

$$y = x^{{1\over x}}$$ Taking logarithm of both sides:

$$\ln y = \frac{\ln x}{x}$$

Differentiating with respect to $x$: $$\frac{1}{y}\frac{dy}{dx}=\frac{1- \ln x}{x^2}$$

$$\frac{dy}{dx}=\frac{x^{{1\over x}}}{x^2}{(1-\ln x)}$$ $$\frac{dy}{dx}=0$$ $\implies $ $\ln x = 1 $ $ \implies x=e$

At $e$: $$\frac{d^2y}{dx^2} = \frac{e^{{1\over e}}}{e^2}\left(\frac{-1}{e}\right) \lt 0 $$

$y$ is maximum at $x=e$

But $x=e$ is the only extreme value , hence $y$ is greatest at $ x=e$

$$e^{{1\over e}} \ge x^{{1\over x}} \ , \forall x \gt 0 $$ $$e^{{1\over e}} \gt \pi^{{1\over \pi}}$$ Raising both side to power $e\pi$ we have

$$\implies e^\pi \gt \pi^e$$