Suppose $\kappa$ is measurable. In trying to show that there are many weakly compact cardinals below it, I quickly reduced the problem to showing that $\{\alpha < \kappa : \alpha$ is weakly compact $\} \in U$, where $U$ is the $\kappa$-complete, nonprincipal ultrafilter on $\kappa$. Then I just need to show that $\kappa$ is weakly compact in the transitive collapse of the ultrapower.
This seems promising to me, but I can't see how to proceed. How can I prove this?
I think the easiest way is to go through the tree property, since $\kappa$ remains strongly inaccessible, this implies weak compactness.
Since the ultrapower is closed under $\kappa$-sequences, if $T$ is a $\kappa$-tree in $M$, it is a $\kappa$-tree in $V$, it has a branch there, and by closure the branch is in $M$.