Assume GCH and let $k,m$ be infinite cardinals. I would like to show that $k^m = \max \{ k,2^m \}$. We of course have $k=\beth_a$ and $m=\beth_b$ for ordinals $a,b$. If $a$ is a successor ordinal, then we may write
$$k^m = (2^{\beth_c})^{\beth_b} = 2^{\beth_c \cdot \beth_b} = 2^{\max \{ \beth_c, \beth_b \}} = \max \{ k,2^m \}$$
where $c$ is the predecessor of $a$. Can we do this when $a$ is not a successor ordinal? If so, how? Also, is GCH required here?
This is not true for singular cardinals. Especially under $\sf GCH$.
If $\kappa$ is a singular cardinal, and $\mu=\operatorname{cf}(\kappa)$ then $\mu<2^\mu=\mu^+<\kappa$ and $\kappa^\mu=\kappa^+$ (due to Koenig's theorem which tells us that $\kappa^{\operatorname{cf}(\kappa)}>\kappa$ for any infinite cardinal) which equals to neither $\kappa$ nor $2^\mu$.
There are many things to say about what we can prove, and what sort of $\sf GCH$ and similar assumptions are needed (more or less) for these things. You can find a lot of that in: