Given a cardinal number $\aleph\_\alpha$, is it known whether or not $\aleph\_\alpha=\beth\_\alpha$ is independent of ZFC?


One could define $\mathrm{CH}(\aleph\_\alpha)$ as $\aleph\_\alpha=\beth\_\alpha$. For which cardinals is it true that $\mathrm{ZFC}\models\mathrm{CH(\kappa)}$?


Clearly, $\mathrm{ZFC}\models\mathrm{CH}(\aleph\_0)$, and $\mathrm{CH}(\aleph\_1)$ is equivalent to $\mathrm{CH}$ (and is thus independent of ZFC). Because $\mathrm{GCH}$ is indepenedent of ZFC, $\neg\mathrm{CH}(\kappa)$ cannot be proven for any cardinal $\kappa$. 


If $\mathrm{ZFC}\models\mathrm{CH}(\aleph\_{\alpha+1})$, then $\aleph\_{\alpha+1}=\beth\_{\alpha+1}$. Thus, if we assume $\aleph\_\alpha<\beth\_\alpha$, we get a contradiction, because $\aleph\_\alpha<\beth\_\alpha$ and then $\aleph\_{\alpha+1}<\beth\_{\alpha+1}$. So, if $\mathrm{CH}(\kappa^+)$ is provable, then $\mathrm{CH}(\kappa)$ is also provable. Thus, for cardinals $\kappa<\aleph\_\omega$, $\mathrm{CH}(\kappa)$ is independent of ZFC.


It is known that $\alpha\leq\aleph\_\alpha$, and thus if $\beth\_\kappa=\kappa$ then $\aleph\_\kappa=\kappa$ (i.e. all $\beth$-fixed points are also $\aleph$-fixed points). Therefore, $\beth\_\kappa=\aleph\_\kappa=\kappa$, and $\beth\_\kappa=\aleph\_\kappa$. So, for all $\beth$-fixed points $\kappa$, $\mathrm{CH}(\kappa)$ is provable from ZFC.


Are there any other known $\kappa$ with $\mathrm{CH}(\kappa)$ independent of ZFC? Is $\exists\kappa\neq\aleph\_0(\mathrm{CH}(\kappa))$ independent of ZFC? (If so, $\beth$ fixed points are independent of ZFC as well.)


Edit: Of course $\beth$-fixed points are not independent of ZFC, they do exist by $\alpha\mapsto\beth\_\alpha$ being a normal function. So, ZFC actually proves the existence of cardinals larger than $\aleph\_0$ which fulfill GCH.



Is it known whether or not $\aleph_\alpha=\beth_\alpha$ can be proven by ZFC?
*Claim.* For any $\alpha >0 \colon$ $\mathrm{ZFC} \not \vdash \mathrm{CH}(\aleph\_{\alpha})$


*Proof.* Starting with $L$, we may enlarge the continuum $\beth\_1$ arbitrarily without changing cardinals. In particular, we may fix a generic $g$ such that $L$ and $L[g]$ have the same cardinals (thus $\aleph\_{\alpha}^L = \aleph\_{\alpha}^{L[g]}$) but $\beth\_1^{L[g]} > \aleph\_{\alpha}^{L[g]}$.


