 Solution

Let $U$ be a nonprincipal $\kappa$-complete ultrafilter on the measurable cardinal $\kappa$, and fix $\lambda <\kappa$. Suppose for a contradiction that $2^{\lambda }\ge \kappa$. Choose distinct subsets $A_{\alpha }\subseteq \lambda$ for $\alpha <\kappa$. For each $\xi <\lambda$, the ultrafilter property chooses exactly one ofas a member $H_{\xi }$ of $U$. Since $\lambda <\kappa$, $\kappa$-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 $|H|\le 1$. 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.