Does $\kappa^\lambda=\kappa$ imply $\mu^\lambda=\mu$ for all $\mu>\kappa$, given $\kappa$, $\mu$, $\lambda$ are cardinals?

Question

This problem is originated from the experience that I was trying to prove 5.19 and 5.20 on Page 60 of Set Theory, Jech(2006).

It seems to be right with AC, but I don't know how to prove it.

Answer 1

No, it does not: if $\mu>\kappa$ has cofinality to $\lambda$, then $\mu^\lambda>\mu$ by König’s theorem (which does require the axiom of choice).

Answer 2

No, not even closely.

Let $\kappa$ be of countable cofinality and $\kappa>\frak c$. We have that ${\frak c}^{\aleph_0}=\frak c$, but $\kappa^{\aleph_0}=\kappa^{\operatorname{cf}(\kappa)}>\kappa$.