= Measurable cardinal is a strong limit cardinal
Let $U$ be a nonprincipal $\kappa$-complete ultrafilter on $\kappa$. If $\lambda<\kappa$ and $(A_\alpha)_{\alpha<\kappa}$ were distinct subsets of $\lambda$, then for every $\xi<\lambda$ choose the $U$-large side of the partition according to whether $\xi\in A_\alpha$. Their intersection is $U$-large by $\kappa$-completeness, but all its indices label the same subset of $\lambda$, so it has at most one member, contradicting nonprincipality. Thus $2^\lambda<\kappa$.
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