= 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 every <limit ordinal> $\lambda$.
We prove simultaneously by <transfinite induction> that every $V_\alpha$ is a <transitive set> and that $V_\alpha\subseteq V_{\alpha+1}$. The claims are immediate for $V_0$. Suppose first that $V_\alpha$ is transitive. If $x\in y\in V_{\alpha+1}=\mathcal P(V_\alpha)$, then $y\subseteq V_\alpha$, so $x\in V_\alpha$. Moreover, transitivity says that every $x\in V_\alpha$ satisfies $x\subseteq V_\alpha$, and hence
$$
x\in\mathcal P(V_\alpha)=V_{\alpha+1}.
$$
Thus $V_{\alpha+1}$ is transitive and $V_\alpha\subseteq V_{\alpha+1}$. At a limit $\lambda$, if $x\in y\in V_\lambda$, then $y\in V_\beta$ for some $\beta<\lambda$; transitivity of $V_\beta$ gives $x\in V_\beta\subseteq V_\lambda$. The inclusions at successor stages also imply
$$
\boxed{V_\alpha\subseteq V_\beta\quad\text{whenever }\alpha\leq\beta.}
$$
Finally, the <rank of a set> $x$ is the <ordinal>
$$
\operatorname{rank}(x)=\sup_{y\in x}\bigl(\operatorname{rank}(y)+1\bigr).
$$
Every $y\in x$ belongs to $V_{\operatorname{rank}(y)+1}\subseteq V_{\operatorname{rank}(x)}$, so $x\subseteq V_{\operatorname{rank}(x)}$. Therefore
$$
\boxed{x\in V_{\operatorname{rank}(x)+1},}
$$
which proves that every set occurs in the hierarchy.
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