Let $\kappa$ be measurable, witnessed by a $\kappa$-complete non-principal ultrafilter $\mathcal{U}$ on $\kappa$.

Step 1: $\kappa$ is regular.

Suppose $\mathrm{cf}(\kappa) = \lambda < \kappa$. Then $\kappa = \bigcup_{\alpha < \lambda} A_\alpha$ for some strictly increasing sequence of ordinals. By $\kappa$-completeness (and $\lambda < \kappa$), the union of $\lambda$ sets in the dual ideal is in the dual ideal: $\bigcup_{\alpha < \lambda} (\kappa \setminus A_\alpha)$ would need to relate properly.

More directly: partition $\kappa = \bigcup_{\alpha < \lambda} B_\alpha$ where $B_\alpha = [\gamma_\alpha, \gamma_{\alpha+1})$ for some cofinal sequence $\gamma_\alpha$. Since $\mathcal{U}$ is $\kappa$-complete and $\lambda < \kappa$, and $\kappa = \bigcup_{\alpha < \lambda} B_\alpha$, we need some $B_\alpha \in \mathcal{U}$. But $|B_\alpha| < \kappa$, and $\mathcal{U}$ is non-principal, so... Actually, for a non-principal $\kappa$-complete ultrafilter, any set of size $< \kappa$ is NOT in $\mathcal{U}$ (since it can be partitioned into singletons, none in $\mathcal{U}$).

Formally: if $|S| < \kappa$ and $S \in \mathcal{U}$, write $S = \bigcup_{s \in S}\{s\}$. By $\kappa$-completeness (since $|S| < \kappa$), some $\{s\} \in \mathcal{U}$, contradicting non-principality. So no set of size $< \kappa$ is in $\mathcal{U}$.

Now if $\mathrm{cf}(\kappa) = \lambda < \kappa$, write $\kappa = \bigcup_{\alpha < \lambda} B_\alpha$ with $|B_\alpha| < \kappa$. Each $B_\alpha \notin \mathcal{U}$, so $\kappa \setminus B_\alpha \in \mathcal{U}$. By $\kappa$-completeness ($\lambda < \kappa$): $\bigcap_{\alpha < \lambda}(\kappa \setminus B_\alpha) = \kappa \setminus \bigcup_\alpha B_\alpha = \emptyset \in \mathcal{U}$. But $\emptyset \notin \mathcal{U}$. Contradiction.

Step 2: $\kappa$ is a strong limit.

Suppose $2^\lambda \geq \kappa$ for some $\lambda < \kappa$. Then there exist $\kappa$ many distinct functions $f_\alpha: \lambda \to 2$ for $\alpha < \kappa$.

For each $\gamma < \lambda$ and $i \in \{0, 1\}$, define $S_{\gamma,i} = \{\alpha < \kappa : f_\alpha(\gamma) = i\}$. Since $\mathcal{U}$ is an ultrafilter, exactly one of $S_{\gamma,0}, S_{\gamma,1}$ is in $\mathcal{U}$. Choose $i_\gamma$ so that $S_{\gamma, i_\gamma} \in \mathcal{U}$.

By $\kappa$-completeness ($\lambda < \kappa$): $T = \bigcap_{\gamma < \lambda} S_{\gamma, i_\gamma} \in \mathcal{U}$. For $\alpha, \beta \in T$: $f_\alpha(\gamma) = i_\gamma = f_\beta(\gamma)$ for all $\gamma < \lambda$, so $f_\alpha = f_\beta$. Thus $|T| = 1$, meaning $T$ is a singleton, contradicting $T \in \mathcal{U}$ (non-principal). $\blacksquare$