Is there a model of ZFC such that $\aleph_\omega^{\aleph_0}=\aleph_{\omega+2}$?

Question

By Easton's theorem, it is consistent with ZFC that $2^{\aleph_0}=\aleph_{\omega+1}$ holds, and if we assume $2^{\aleph_0}=\aleph_{\omega+1}$ then $$\aleph_{\omega+1}\le 2^{\aleph_0}\le \aleph_\omega^{\aleph_0}\le \aleph_{\omega+1}^{\aleph_0} =(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}=\aleph_{\omega+1}$$ So it is consistent with ZFC that $\aleph_\omega^{\aleph_0}=\aleph_{\omega+1}$, and I conjecture that $\aleph_\omega^{\aleph_0}=\aleph_{\omega+2}$ is consistent with ZFC. But how to prove it? Thanks for any help.

Answer 1

Yes. Easily (if one is familiar with forcing). Start with a model of $\sf GCH$. Now force that for every $\alpha\in\omega\cup\{\omega+1\}$:

$$2^{\aleph_\alpha}=\aleph_{\omega+2}$$

Then we have the wanted equality, by the same considerations as in your question.

(In fact, if one starts with $\sf GCH$ then we only need to force $2^{\aleph_0}=\aleph_{\omega+2}$, the rest follows.)