[proofplan] A strongly inaccessible cardinal is, by definition, an uncountable regular strong limit cardinal. Since weak inaccessibility already gives that $\kappa$ is uncountable and regular, it remains only to prove the strong limit condition. We fix a cardinal $\lambda < \kappa$ and show $2^\lambda < \kappa$, treating finite $\lambda$ directly and infinite $\lambda$ by GCH. The limit-cardinal hypothesis prevents the successor cardinal $\lambda^+$ from being equal to $\kappa$. [/proofplan] [step:Reduce the conclusion to proving the strong limit condition] Since $\kappa$ is weakly inaccessible, $\kappa$ is an uncountable regular limit cardinal. To prove that $\kappa$ is strongly inaccessible, it is therefore enough to prove that $\kappa$ is a strong limit cardinal, meaning that for every cardinal $\lambda < \kappa$ one has $2^\lambda < \kappa$. [/step] [step:Handle finite cardinals below $\kappa$] Let $\lambda < \kappa$ be a finite cardinal. Then $2^\lambda$ is also finite. Since $\kappa$ is uncountable, every finite cardinal is strictly below $\kappa$, and hence $2^\lambda < \kappa$. [/step] [step:Use GCH and the limit-cardinal property for infinite cardinals below $\kappa$] Let $\lambda < \kappa$ be an infinite cardinal. By GCH, applied to the infinite cardinal $\lambda$, we have \begin{align*} 2^\lambda = \lambda^+. \end{align*} Because $\lambda < \kappa$ and $\kappa$ is a cardinal, the definition of the successor cardinal $\lambda^+$ as the least cardinal strictly larger than $\lambda$ gives \begin{align*} \lambda^+ \leq \kappa. \end{align*} If $\lambda^+ = \kappa$, then $\kappa$ would be a successor cardinal, contradicting that $\kappa$ is a limit cardinal. Therefore $\lambda^+ < \kappa$, and hence \begin{align*} 2^\lambda < \kappa. \end{align*} [guided] Fix an infinite cardinal $\lambda < \kappa$. The goal is to prove $2^\lambda < \kappa$. Since $\lambda$ is infinite, the Generalized Continuum Hypothesis applies to $\lambda$ and says exactly that the cardinality of the power set of a set of size $\lambda$ is the next cardinal after $\lambda$: \begin{align*} 2^\lambda = \lambda^+. \end{align*} It remains to show that this successor cardinal is still below $\kappa$. By definition, $\lambda^+$ is the least cardinal strictly greater than $\lambda$. Since $\kappa$ is itself a cardinal and $\lambda < \kappa$, this leastness gives \begin{align*} \lambda^+ \leq \kappa. \end{align*} There are now only two possibilities: either $\lambda^+ < \kappa$ or $\lambda^+ = \kappa$. The second possibility would say that $\kappa$ is the successor of the smaller cardinal $\lambda$. That is incompatible with the hypothesis that $\kappa$ is a limit cardinal, since a limit cardinal is not a successor cardinal. Therefore \begin{align*} \lambda^+ < \kappa. \end{align*} Combining this with the GCH identity $2^\lambda = \lambda^+$ gives \begin{align*} 2^\lambda < \kappa. \end{align*} [/guided] [/step] [step:Conclude that $\kappa$ is strongly inaccessible] The preceding two steps show that for every cardinal $\lambda < \kappa$, whether finite or infinite, one has $2^\lambda < \kappa$. Thus $\kappa$ is a strong limit cardinal. Since $\kappa$ is already uncountable and regular by weak inaccessibility, $\kappa$ is an uncountable regular strong limit cardinal. Therefore $\kappa$ is strongly inaccessible. [/step]