 Solution

Every finite ordinal belongs to $V_{\omega }$, so $\omega \subseteq V_{\omega }$ and $\omega \in \mathcal{P}(V_{\omega })=V_{\omega +1}$. However, $\omega \notin V_{\omega }$, whose elements are the hereditarily finite sets.

Suppose $y\in V_{\omega +1}$ were an internal power set of $\omega$. Internally $\omega \subseteq \omega$, and $\omega$ is an available element of the structure. Thus the Axiom of power set would require $\omega \in y$. But $y\in \mathcal{P}(V_{\omega })$ means $y\subseteq V_{\omega }$, which would put $\omega \in V_{\omega }$, a contradiction. ThereforeThis proves power-set failure at the first rank above V omega by an internal obstruction, including the subset $\omega$ itself.