= Solution
Two facts about a <transitive set> $A$ drive the <transfinite induction>. First, $A\subseteq\mathcal P(A)$: every $x\in A$ is a subset of $A$. Second, $\mathcal P(A)$ is a <transitive set>: if $x\in y\in\mathcal P(A)$, then $x\in A$, hence $x\subseteq A$, hence $x\in\mathcal P(A)$.
Starting with $V_0=\varnothing$, these facts prove the successor step. At a <limit ordinal> $\lambda$, the union of the earlier increasing <transitive sets> is transitive: if $x\in y\in V_\lambda$, some $\alpha<\lambda$ contains $y$ in $V_\alpha$, and transitivity puts $x$ there too. The union contains every earlier stage. Thus the induction gives both transitivity and $V_\beta\subseteq V_\alpha$ whenever $\beta\leq\alpha$.
Finally $V_\alpha\subseteq V_\alpha$, so $V_\alpha\in\mathcal P(V_\alpha)=V_{\alpha+1}$. The <Axiom of foundation> gives $V_\alpha\notin V_\alpha$. This witnesses strictness:
$$
\boxed{V_\alpha\text{ is transitive},\qquad V_\alpha\subsetneq V_{\alpha+1}.}
$$
This is the <transitivity and strict growth of the cumulative hierarchy>.
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