TITLE: some basic cardinal arithmetic on $\text{cf}(\aleph_{\omega_1})$
QUESTION [5 upvotes]: I'm reading The Joy of Sets by K. Devlin, by self-study. I've just seen a statement $\text{cf}(\aleph_{\omega_1})=\omega_1$ without proof, but I think this is slightly harder to prove than more obvious one, $\text{cf}(\aleph_{\omega})=\omega$. 
Concretely, $\text{cf}(\aleph_{\omega_1})\le \omega_1$ is trivial, but.. how about the other direction? I've tried and I guess $\aleph_{\omega_1}^{\aleph_0}=\aleph_{\omega_1} $ (as cardinal exp.) implies the result desired. How can I prove the last identity, or is there a simpler proof on the cofinality?

REPLY [3 votes]: The actual question has been answered, but it should still be pointed out that $\aleph_{\omega_1}^{\aleph_0}=\aleph_{\omega_1}$ is not necessarily true, or rather it is not provable without additional assumptions. 
For example, it is possible that $2^{\aleph_0}=\aleph_{\omega_2}$, in which case it is trivial that $\aleph_{\omega_1}^{\aleph_0}\geq\aleph_{\omega_2}$.