[proofplan] The definition of a Mahlo cardinal gives regularity, uncountability, and stationarily many strongly inaccessible cardinals below $\kappa$. Since every stationary subset of a regular uncountable cardinal is unbounded, below any cardinal $\lambda < \kappa$ we can find a strongly inaccessible cardinal $\delta$ with $\lambda < \delta < \kappa$. The strong limit property of $\delta$ then gives $2^\lambda < \delta < \kappa$, proving that $\kappa$ is a strong limit; together with regular uncountability, this is exactly strong inaccessibility. [/proofplan] [step:Use Mahloness to obtain an unbounded supply of inaccessible cardinals below $\kappa$] Let $S$ denote the class of strongly inaccessible cardinals below $\kappa$: \begin{align*} S := \{\delta < \kappa : \delta \text{ is strongly inaccessible}\}. \end{align*} By the definition of a Mahlo cardinal, $\kappa$ is regular and uncountable, and $S$ is stationary in $\kappa$. Since stationary subsets of regular uncountable cardinals are unbounded in the ambient cardinal, $S$ is unbounded in $\kappa$ (citing a result not yet in the wiki: Stationary subsets of regular uncountable cardinals are unbounded). Thus, for every ordinal $\alpha < \kappa$, there exists $\delta \in S$ such that \begin{align*} \alpha < \delta < \kappa. \end{align*} [guided] We isolate the part of the Mahlo hypothesis that will be used to prove the strong limit property. Define \begin{align*} S := \{\delta < \kappa : \delta \text{ is strongly inaccessible}\}. \end{align*} This notation means that $S$ contains exactly those cardinals below $\kappa$ which are already strongly inaccessible. By the stated definition of a Mahlo cardinal, two facts hold: first, $\kappa$ is regular and uncountable; second, the set $S$ is stationary in $\kappa$. The point of stationarity is that it is stronger than unboundedness when the ambient cardinal is regular and uncountable. The theorem we use is: every stationary subset of a regular uncountable cardinal is unbounded in that cardinal (citing a result not yet in the wiki: Stationary subsets of regular uncountable cardinals are unbounded). Its hypotheses are satisfied because the Mahlo assumption gives that $\kappa$ is regular and uncountable, and that $S$ is stationary in $\kappa$. Therefore $S$ is unbounded in $\kappa$. Unboundedness of $S$ in $\kappa$ means exactly that, for every ordinal $\alpha < \kappa$, there is some $\delta \in S$ above $\alpha$ but still below $\kappa$: \begin{align*} \alpha < \delta < \kappa. \end{align*} This is the mechanism that will let us dominate each power set cardinal $2^\lambda$ by a smaller inaccessible cardinal $\delta$ below $\kappa$. [/guided] [/step] [step:Prove that $\kappa$ is a strong limit cardinal] Let $\lambda$ be an arbitrary cardinal with $\lambda < \kappa$. Since $S$ is unbounded in $\kappa$, choose $\delta \in S$ such that \begin{align*} \lambda < \delta < \kappa. \end{align*} Because $\delta \in S$, the cardinal $\delta$ is strongly inaccessible. In particular, $\delta$ is a strong limit cardinal, so for every cardinal $\mu < \delta$ one has $2^\mu < \delta$. Applying this with $\mu = \lambda$ gives \begin{align*} 2^\lambda < \delta. \end{align*} Since $\delta < \kappa$, we obtain \begin{align*} 2^\lambda < \kappa. \end{align*} Because $\lambda < \kappa$ was arbitrary, $\kappa$ is a strong limit cardinal. [/step] [step:Combine strong limitness with regular uncountability] The Mahlo hypothesis already gives that $\kappa$ is regular and uncountable. The previous step proves that $\kappa$ is a strong limit cardinal. By the definition of a strongly inaccessible cardinal, a cardinal is strongly inaccessible precisely when it is uncountable, regular, and strong limit. Hence $\kappa$ is strongly inaccessible. [/step]