= Solution
Let $U$ witness that $\kappa$ is <measurable cardinal>[measurable]. Regularity is given, so it remains to prove the <strong limit cardinal> property. First, every $A\in U$ has cardinality $\kappa$: if $|A|<\kappa$, then
$$
\kappa\setminus A=\bigcap_{\alpha\in A}(\kappa\setminus\{\alpha\})\in U
$$
by nonprincipality and $\kappa$-completeness, contradicting $A\in U$.
Suppose $\lambda<\kappa$ and $2^\lambda\geq\kappa$. Choose an injection $f:\kappa\to\mathcal P(\lambda)$. For each $\xi<\lambda$, exactly one of
$$
A_\xi=\{\alpha<\kappa:\xi\in f(\alpha)\},
\qquad \kappa\setminus A_\xi
$$
lies in $U$. Their chosen intersection lies in $U$ by $\kappa$-completeness. On that intersection every $f(\alpha)$ is the same subset of $\lambda$, contradicting injectivity because every member of $U$ has size $\kappa$. Thus $2^\lambda<\kappa$, and $\kappa$ is strongly inaccessible.
Back to article page
