Is the following equality independent of ZFC: $$\bigcup_{\aleph_{\beta}<\beth_{\omega}}2^{\aleph_{\beta}}=\beth_\omega $$
Consistent: Now that the equality is consistent with ZFC since it holds assuming GCH. In ZFC+GCH we have that $$\bigcup_{\aleph_{\beta}<\beth_{\omega}}2^{\aleph_{\beta}}=\bigcup_{\aleph_{\beta}<\aleph_{\omega}}2^{\aleph_{\beta}}=\bigcup_{ \aleph_\beta<\aleph_\omega}\aleph_{\beta+1}=\bigcup_{\aleph_\beta<\aleph_\omega}\aleph_\beta=\aleph_\omega =\beth_\omega$$
Note that it is easy to show that the argument above can be expanded to ZFC+ "$\aleph_\omega$ is a strong limit cardinal" pretty easily.
I'm just not sure how to prove this equality (if it is provable) in the general case.
Thanks!
No, equality is quite provable.
Note that if $\kappa<\beth_\omega$ then for some $n\in\omega$ we have that $\kappa<\beth_n$, therefore $2^\kappa\leq 2^{\beth_n}=\beth_{n+1}$. Therefore it is always true that $$\beth_\omega=\sup\{2^\kappa\mid\kappa<\beth_\omega\}.$$
In other words, $\beth_\omega$ is provably a strong limit cardinal.