Prove that $|V_\alpha|=|\operatorname{P}(\alpha)|$ if and only if $\alpha=\{2,\omega+1\}$ or $\alpha=\kappa+1$, $\kappa=\beth_\kappa$

$$\kappa$$ is a cardinal, $$V_\alpha$$ belongs to the Von Neumann hierarchy $$\begin{cases} V_0=\emptyset \\ V_{\alpha+1}=P(V_\alpha) \\ V_\lambda=\underset{\gamma<\lambda}{\bigcup}V_\gamma \end{cases}$$ and the Beth function is defined in this way: $$\begin{cases} \beth_0=\aleph_0 \\ \beth_{\alpha+1}=2^{\beth_\alpha} \\ \beth_\lambda=\underset{\gamma<\lambda}{\bigcup}\beth\gamma \end{cases}$$

It's easy to see that $$|V_0|\ne|\operatorname{P}(0)|, \; |V_1|\ne|\operatorname{P}(1)|, \; |V_2|=|\operatorname{P}(2)|$$ and, for countable recursion, I prooved that $$\forall n\in\omega \; |V_n|>|\operatorname{P}(n)|$$.

$$V_\omega$$ is countable, whereas $$|V_{\omega+1}|=2^{|V_\omega|} =2^{\aleph_0}=|\operatorname{P}(\omega+1)|.$$ Then, $$\forall \; \omega+2<\alpha<\omega^2 \quad |V_\alpha|>2^{\aleph_0}=|\operatorname{P}(\alpha)|$$ because these $$\alpha$$ are countable.

Now, for ordinals $$\alpha\geq\omega^2$$ I use this fact: $$|V_\alpha|=\beth_\alpha$$. Let be $$\kappa$$ a cardinal, $$\forall\alpha+2$$ such that $$|\alpha|=\kappa$$, then $$|V_{\alpha+2}|=\beth_{\alpha+2}=2^{\beth_{\alpha+1}}>\beth_{\alpha+1}=2^{\beth_{\alpha}}\geq2^{|\alpha|}\geq2^{\kappa}=\operatorname{P}(\alpha+2)$$.

Cardinals and successor of cardinals are left. $$\forall\kappa$$ cardinal $$|V_\kappa|=\sum_{\gamma<\kappa}{|V_\gamma|}=\max\{\sup_{\gamma<\kappa}{|V_\gamma|,\kappa}\}$$ and I don't know how to show that it isn't equal to $$|\operatorname{P}(\kappa)|.$$ If $$\kappa$$ is a fixed point of Beth function, then $$|V_{\kappa+1}|=|\operatorname{P}(\kappa+1)|$$, if $$\kappa$$ isn't a fixed point, it shouldn't be true, but I don't know how to go on.

Suppose $$\kappa$$ is a cardinal and $$\beth_\kappa >\kappa$$. Then by definition of $$\beth_\kappa$$ (since $$\kappa$$ is a limit ordinal), $$\beth_\gamma >\kappa$$ for some $$\gamma < \kappa$$. This should tell you that $$\beth_\kappa > 2^\kappa$$ and so you should be done.
Same thing for $$\kappa +1$$ : if $$\kappa$$ isn't a fixed point, you still have that $$\beth_\gamma >\kappa$$ for some $$\gamma<\kappa$$, etc.