= Solution
The <Von Neumann hierarchy> is defined by <transfinite recursion>:
$$
V_0=\varnothing,
\qquad
V_{\alpha+1}=\mathcal P(V_\alpha),
\qquad
V_\lambda=\bigcup_{\beta<\lambda}V_\beta
$$
for a limit ordinal $\lambda$.
We prove by <transfinite induction> that every $V_\alpha$ is a <transitive set>. The claim is immediate for $V_0$. If $V_\alpha$ is transitive and $x\in y\in V_{\alpha+1}$, then $y\subseteq V_\alpha$, so $x\in V_\alpha$. Transitivity also gives $x\subseteq V_\alpha$, hence $x\in\mathcal P(V_\alpha)=V_{\alpha+1}$. At a limit stage, if $x\in y\in V_\lambda$, then $y\in V_\beta$ for some $\beta<\lambda$, so $x\in V_\beta\subseteq V_\lambda$.
Transitivity implies $V_\alpha\subseteq\mathcal P(V_\alpha)=V_{\alpha+1}$. Induction and unions at limit stages then give
$$
\boxed{\alpha\leq\beta\Longrightarrow V_\alpha\subseteq V_\beta}.
$$
By well-founded recursion on membership, define the <rank of a set>
$$
\operatorname{rank}(x)
=\sup_{y\in x}\bigl(\operatorname{rank}(y)+1\bigr).
$$
Induction on this rank gives $y\in V_{\operatorname{rank}(x)}$ for every $y\in x$. Thus
$$
x\subseteq V_{\operatorname{rank}(x)}
$$
and consequently
$$
\boxed{x\in V_{\operatorname{rank}(x)+1}}.
$$
Every set therefore occurs at some level of the hierarchy.
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