Apparently, for all ordinals $\alpha > \omega$, the following two are equivalent:
- $|L_{\alpha}| = |V_{\alpha}|$
- $\alpha = \beth_{\alpha}$
Where $L$ is the constructible universe and $V$ the von Neumann universe and $\beth_{\alpha}$ is the Beth sequence indexed on $\alpha$ (the Beth sequence is defined by $\beth_0 = \aleph_0$; $\beth_{\alpha + 1} = 2^{\beth_\alpha}$ and $\beth_\lambda = \bigcup_{\alpha < \lambda} \beth_{\alpha}$).
We know that if $\alpha \geq \omega$, then $|L_{\alpha}| = |\alpha|$. I thought about showing that $|V_{\alpha}| = \beth_{\alpha}$ and we know that $\beth_{\alpha}$ is a cardinal, so then $|L_{\alpha}| = \alpha$ and we have the equivalence.
But how can I show that $|V_{\alpha}| = \beth_{\alpha}$? I thought about induction, but it is very hard to find a base case.
The proof that $|V_\alpha|=\beth_\alpha$ is in fact not true. $|V_{\omega+\alpha}|=\beth_\alpha$. However, after $\alpha=\omega^2$, you get that $\omega+\alpha=\alpha$. So this must certainly be the case for a $\beth$-fixed point cardinal.
Now you can use induction to finish your proof quite easily.