I have already know how to prove $\pi^{e} < e^{\pi}$(solution), but I cannot figure out how to prove $\pi^{e} + 1 > e^{\pi}$. You can use approximation to prove it but calculators are prohibited. Any help is appreciated, thanks!
I have tried to convert $\pi^{e}$ to $e^{e\ln\pi}$ but I couldn't think of what to do next.
Just some numerical thoughts, take this as a long comment. I will add something later!
If we think about it as the following inequality
$$x^e > e^x -1$$ Then all we need is to find the range of $x$ for which that inequality is true. This reduces to study
$$\dfrac{\ln(e^x-1)}{\ln(x)} - e < 0$$
If $\pi$ lies in that range, we are done! One interval for which this holds is surely $x\in(0,1)$. The other one can be found numerically and it's very approximatively $x \in (1.87, 3.22) \ni \pi$
Clearly this is not a number theory proof, nor an analytical proof. Nor a rigorous one, but it might be a starting point. Will add more later!