= 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 <Axiom of power set>[power-set axiom] holds in $V_\lambda$.
Here the absoluteness fact used is: every <bounded formula in set theory>[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 \b[<Infinity> holds as well]. In contrast $V_\omega$ has only hereditarily finite <sets> and cannot use $\omega$ as an element.
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