 Solution

Take a nonzero limit ordinal $\lambda$ and $x\in V_{\lambda }$. Choose $\beta <\lambda$ with $x\in V_{\beta }$. Transitivity gives $x\subseteq V_{\beta }$. Hence every actual subset of $x$ belongs to $\mathcal{P}(V_{\beta })=V_{\beta +1}$, and the actual set $\mathcal{P}(x)$ belongs to $V_{\beta +2}$. Since $\lambda$ is a limit, $\beta +2<\lambda$, so $\mathcal{P}(x)\in V_{\lambda }$. Thus the internal power set contains every actual subset of $x$, and the power-set axiom holds in $V_{\lambda }$.

Here the absoluteness fact used is: every bounded formula is absolute between transitive sets or classes with the actual membership relation. In particular $z\subseteq x$, expressed by $\forall t\in z(t\in x)$, has the same meaning internally and externally. The argument supplies the actual power set as an internal witness. If zero is included in one's convention for limit ordinals, its empty stage has no nonempty first-order domain; the assertion is understood for nonzero limits.

If also $\lambda >\omega$, the preceding ordinal-membership result gives $\omega \in V_{\omega +1}\subseteq V_{\lambda }$. The actual set $\omega$ contains the empty set and is closed under $z\mapsto z\cup \{z\}$. These bounded membership and successor conditions are absolute in the transitive stage, so $\omega$ is an internal inductive set. Therefore Infinity holds as well. In contrast $V_{\omega }$ has only hereditarily finite sets and cannot use $\omega$ as an element.