some basic cardinal arithmetic on $\text{cf}(\aleph_{\omega_1})$ 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?

Suppose that $\langle\alpha_n:n\in\omega\rangle$ is an increasing sequence cofinal in $\omega_{\omega_1}$. For each $\xi<\omega_1$ there is a least $n(\xi)\in\omega$ such that $\omega_\xi\le\alpha_{n(\xi)}$. Clearly there is then an $m\in\omega$ such $X=\{\xi<\omega_1:n(\xi)=m\}$ is uncountable, and it follows easily that $\omega_{\omega_1}=\sup_{\xi\in X}\omega_\xi\le\alpha_m<\omega_{\omega_1}$, which is absurd.
