 Solution

We prove the stronger ordinal identityby transfinite induction. It holds at zero. At a successor stage an ordinal $\gamma$ belongs to $\mathcal{P}(V_{\beta })$ exactly when every $\delta <\gamma$ belongs to $V_{\beta }$. By induction this means every $\delta <\gamma$ is less than $\beta$, equivalently $\gamma \le \beta$. Thus the ordinals in $V_{\beta +1}$ are precisely those less than $\beta +1$. At a limit the identity follows by taking unions.

Every $\beta <\alpha$ consequently belongs to $V_{\alpha }$, so $\alpha \subseteq V_{\alpha }$ and $\alpha \in \mathcal{P}(V_{\alpha })=V_{\alpha +1}$. But $\alpha \notin V_{\alpha }$, since an ordinal is not less than itself. HenceThe only well-foundedness used is that of the ordinal order, available independently of Foundation for arbitrary ambient sets.