= Solution
Let $U$ be a nonprincipal $\kappa$-complete <ultrafilter> on the measurable cardinal $\kappa$, and fix $\lambda<\kappa$. Suppose for a contradiction that $2^\lambda\geq\kappa$. Choose distinct subsets $A_\alpha\subseteq\lambda$ for $\alpha<\kappa$. For each $\xi<\lambda$, the <ultrafilter> property chooses exactly one of
$$
S_\xi=\{\alpha<\kappa:\xi\in A_\alpha\},
\qquad
\kappa\setminus S_\xi
$$
as a member $H_\xi$ of $U$. Since $\lambda<\kappa$, <kappa-complete filter>[$\kappa$-completeness] gives $H=\bigcap_{\xi<\lambda}H_\xi\in U$.
Any two indices in $H$ give subsets having the same membership decision at every $\xi<\lambda$, so they give the same $A_\alpha$. The chosen subsets were distinct, hence $|H|\leq1$. This contradicts the fact that a <small set is absent from a complete nonprincipal ultrafilter>. Therefore $2^\lambda<\kappa$ for every $\lambda<\kappa$, which is precisely the <strong limit cardinal> condition. This is the <measurable cardinal is a strong limit cardinal> argument.
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