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 $2^{\lambda }<\kappa$.