Let $\lambda$ and $\kappa$ be cardinals. I think the statement
$0 \lt \lambda \lt \operatorname{cof} \kappa \Rightarrow \kappa^{\lambda}=\kappa$
is true in ZFC, but not sure.
Edit: The statement is false in general, as Keith show with the case $\aleph_{0} < \operatorname{cof}(\aleph_{1})$ but $\aleph_{1}=\aleph_{1}^{\aleph_{0}}$ is equivalent to the continuum hypothesis, what is know to be independet of ZFC (to see this remember $2^{\aleph_{0}}\leq \aleph_{1}^{\aleph_{0}}$)
There are isteresting situations where the statement is true, as $\lambda$ finite or $\kappa= \beth_{\beta}$ for $\beta$ a limit ordinal.
This two cases are all the cases?
For the ones who are learning these things, like me: $\beth$ is a cardinal function defined as
$\beth_{0}=\aleph_{0}$
$\beth_{\beta}=\displaystyle\bigcup_{\alpha< \beta}2^{\beth_{\alpha}}$