Proof that $$p^{q^p} > q^{p^q} $$ for $p>q>1$
My idea
I am convinced that it can be transformed into inequality wih one additional variable. But firstly we should transform that.
$\ln$ increases so our thesis is true if and only if:
$$ q^p \ln p > p^q \ln q \\
\ln(q^p \ln p) > \ln(p^q \ln q) \\
p\ln q+\ln \ln p > q\ln p + \ln \ln q$$
But there I don't know how to transofrm it further. If I could reduce that to 1 variable, then I can make a function and in use of derivatives can solve that problem.
By your work we need to prove that $$p\ln{q}-q\ln{p}+\ln\ln{p}-\ln\ln{q}>0$$ or $$\ln\frac{\ln{p}}{\ln{q}}>q\ln{p}-p\ln{q},$$ which is obvious for $q\ln{p}-p\ln{q}\leq0.$
Thus, it's enough to prove our inequality for $$q\ln{p}>p\ln{q}$$ or $$\ln{q}+\ln\ln{p}>\ln{p}+\ln\ln{q}$$ or $$\ln\frac{\ln{p}}{\ln{q}}>\ln{p}-\ln{q}.$$ Id est, it's enough to prove that $$\ln{p}-\ln{q}>q\ln{p}-p\ln{q}$$ or $$\frac{p-1}{\ln{p}}>\frac{q-1}{\ln{q}},$$ which is true because $f(x)=\frac{x-1}{\ln{x}}$ increases.