Consider the cardinals $\kappa:=2^{2^{\aleph_0}}$, $\lambda:=2^\kappa$, $\mu:=2^\lambda$. Which cardinal is larger, $\kappa^{\mu^\lambda}$ or $\lambda^{\mu^\kappa}$?
The only rules I believe I need are $2^c>c$ and $a.b=\max \{a,b\}$ for infinite cardinals $a,b,c$. I see that
$$\mu^\lambda = (2^{2^{2^{2^{\aleph_0}}}})^{2^{2^{2^{\aleph_0}}}} = 2^{(2^{2^{2^{\aleph_0}}}.2^{2^{2^{\aleph_0}}})} = 2^{2^{2^{2^{\aleph_0}}}}$$ and $$ \mu^\kappa = (2^{2^{2^{2^{\aleph_0}}}})^{2^{2^{\aleph_0}}} = 2^{(2^{2^{2^{\aleph_0}}}.2^{2^{\aleph_0}})} = 2^{2^{2^{2^{\aleph_0}}}} = \mu $$
so
$$ \kappa^{\mu^\lambda} = (2^{2^{\aleph_0}})^{2^{2^{2^{2^{\aleph_0}}}}} = 2^{2^{2^{2^{2^{\aleph_0}}}}} $$
and
$$ \lambda^{\mu^\kappa} = {(2^{2^{2^{\aleph_0}}})}^{2^{2^{2^{2^{\aleph_0}}}}} = 2^{2^{2^{2^{2^{\aleph_0}}}}}$$
but then I get that $\kappa^{\mu^\lambda} = \lambda^{\mu^\kappa}$, which means that either I or the question I'm doing is incorrect. Which one is it?
Your solution is correct, here is a more readable way of looking at it, remembering that $\mu>\lambda>\kappa$ by Cantor's theorem to simplify the products:
$$\mu^\lambda=(2^\lambda)^\lambda=2^{\lambda\cdot\lambda}=2^\lambda=\mu$$ $$\mu^\kappa=(2^\lambda)^\kappa=2^{\lambda\cdot\kappa}=2^\lambda=\mu$$
Now we're interested in comparing $\kappa^\mu$ and $\lambda^\mu$, but those are both $2^\mu$, since $\mu>\kappa$ and $\mu>\lambda$ (whenever you have two cardinals $\eta>\xi$, $\xi^\eta=2^\eta$).