How do I prove that if $x \in V_\alpha$, then $\bigcup x \in V_{\alpha + 1}$, where $V_{\alpha+1} = P(V_\alpha)$ and $V$ is the von Neumann Universe? This is a problem that was on a final I took in my Set Theory class and I just couldn’t figure it out.  The test is over and I already got the problem wrong, I’m just trying to go back and figure it out and make sure I understand the material before moving on to the next course.
I have a photo of an attempted solution but have less than 10 reputation.  
The upshot is that I can’t figure out how to get $\bigcup x \in P(V_\alpha)$.  I’ve tried everything I can think of for hours with no luck.  I’ve done everything that I’m ever gonna do at this point, still no luck.   
The best I’ve got is that $x\in V_{\alpha+1}$ since $V_{\alpha}$ is a subset of $V_{\alpha+1}$. But this leaves no clear path into how the union of $x$ is in the power set of $V_{\alpha}$. 
 A: Let $x \in V_{\alpha}$. Recall that
$$
\bigcup x = \{ z \mid \exists y \in x \colon z \in y\} = \bigcup_{y \in x} y.
$$
For any $y \in x$ we have, because $V_{\alpha}$ is transitive, that $y \subseteq V_{\alpha}$. Hence $\bigcup x = \bigcup_{y \in x} y \subseteq V_{\alpha}$ and thus $\bigcup x \in \mathcal P(V_{\alpha}) = V_{\alpha + 1}$.

If you already know that $V_{\alpha} = \{x \mid \mathrm{rank}_{\in}(x) < \alpha\}$, there is an even easier argument: Every element $z \in \bigcup x$ has rank $\mathrm{rank}_{\in}(z) < \mathrm{rank}_{\in}(x) < \alpha$. Since
$$
\mathrm{rank}_{\in}(\bigcup x) = \sup \{ \mathrm{rank}_{\in}(z)+1 \mid z \in \bigcup x \},
$$
we have $\mathrm{rank}_{\in}(\bigcup x) \le \mathrm{rank}_{\in}(x) < \alpha$ and hence $\bigcup x \in V_{\alpha} \subseteq V_{\alpha+1}$. (And this result is optimal since it is possible that $\bigcup x = x$, e.g. when $x$ is a limit ordinal.)
A: From the def'n $V_a=\cup_{b\in a}V_b$ if $a=\cup a,$ and $V_{a+1}=P(V_a)$ we obtain by transfinite induction on $a$ that $V_a$ is a transitive set: $\forall x\in V_a\;(x\subset V_a).$ 
Let $x\in V_a.$  Then for all $y $ we have $$y\in x\in V_a\implies y \in x\subset V_a\implies y\in V_a\implies y\subset V_a.$$   So $\cup x=\cup \{y:y\in x\}$ is a union of subsets of $V_a,$ so $\cup x\in P(V_a)=V_{a+1}.$
