= Solution
Use <transfinite induction>. The inclusion at zero is immediate. If $V_\beta\subseteq V_{\beta+1}$, monotonicity of the <power set> gives
$$
V_{\beta+1}=\mathcal P(V_\beta)\subseteq\mathcal P(V_{\beta+1})=V_{\beta+2}.
$$
If $\alpha$ is a limit and $x\in V_\alpha$, choose $\beta<\alpha$ with $x\in V_\beta$. The inductive inclusion $V_\beta\subseteq\mathcal P(V_\beta)$ means that $x\subseteq V_\beta\subseteq V_\alpha$. Thus $x\in\mathcal P(V_\alpha)=V_{\alpha+1}$. This proves \b[$V_\alpha\subseteq V_{\alpha+1}$ for every <ordinal>]. These successive inclusions and the union definition also give $V_\beta\subseteq V_\alpha$ whenever $\beta\leq\alpha$.
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