$\kappa$ 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.
Best Answer
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.