TITLE: Cardinal Arithmetic: $\beth_\omega$
QUESTION [5 upvotes]: 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!

REPLY [6 votes]: 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.