Cardinal arithmetic- Problem 5.18 Jech's book I am solving problems of the Jech's book (set theory). I need help to solve problem 5.18: $ℵ_ω^{ℵ_1} = ℵ_ω^{ℵ_0} · 2^{ℵ_1}$. I have some computation in the following way, but I think my path is not correct.
Obviously, $\aleph_{\omega}^{{\aleph_{1}}}$ is at
least $\aleph_{\omega}^{{\aleph_{0}}}2^{{\aleph_{1}}}$. Let’s prove the
equality.
By theorem 5.20, it is either ${\aleph_{n}}^{{\aleph_{0}}}$ for
some $n<\omega$ or
(since $\operatorname{cf}(\aleph_{\omega})=\aleph_{0}\leq\aleph_{1})$ it
is $\aleph_{\omega}^{{\aleph_{0}}}$. In the latter case, using Hausdorff
formula we
get ${\aleph_{n}}^{{\aleph_{0}}}=\aleph_{1}^{{\aleph_{1}}}{\aleph_{2}...\aleph_{n}}\leq 2^{{\aleph_{1}}}\aleph_{\omega}^{{\aleph_{0}}}$.
Do you have a solution for this problem?
 A: I will try to fill the gaps in your idea.
As you said, one direction is clear: $\aleph_\omega^{\aleph_0}\leq\aleph_\omega^{\aleph_1}$ and $2^{\aleph_1}\leq\aleph_\omega^{\aleph_1}$, hence $\aleph_\omega^{\aleph_0} \cdot 2^{\aleph_1} = max \{ \aleph_\omega^{\aleph_0}, 2^{\aleph_1}  \}\leq \aleph_\omega^{\aleph_1}$.
For the other direction, we apply Theorem 5.20 in Jech. Firstly, notice that we have that $\aleph_\omega> \aleph_1$ and also $cf(\aleph_\omega)=\aleph_0$. Then we have two cases:
(a) If there is $\mu<\aleph_\omega$ such that $\mu ^{\aleph_1}\geq\aleph_\omega$, then this means that for some $\aleph_n$ we have $\aleph_n^{\aleph_1}\geq\aleph_\omega$. Then by item (ii) in Theorem 5.20 we have that $\aleph_\omega^{\aleph_1}=\aleph_n^{\aleph_1}$. Then, by  applying n times Hausdorff's formula we obtain:
$$\aleph_\omega^{\aleph_1}=\aleph_n^{\aleph_1}= \aleph_n \cdot \aleph_{n-1}\cdot\dots\cdot\aleph_0^{\aleph_1} .   $$
But $\aleph_n \cdot \aleph_{n-1}\cdot\dots\cdot\aleph_0^{\aleph_1}  = max\{\aleph_n , \aleph_{n-1}\dots\aleph_0^{\aleph_1} \}\leq \aleph_\omega^{\aleph_0} \cdot 2^{\aleph_1} $.
(b) Otherwise, suppose for all $\mu<\aleph_\omega$ we have $\mu^{\aleph_1}<\aleph_\omega$, then we can apply item (iiib) in Jech's Theorem 5.20, thereby getting that:
$$\aleph_\omega^{\aleph_1} =   \aleph_\omega^{cf (\aleph_\omega)} = \aleph_\omega ^{\aleph_0}\leq \aleph_\omega^{\aleph_0} \cdot 2^{\aleph_1} .$$
Hence in both case we obtain $\aleph_\omega^{\aleph_1}\leq \aleph_\omega^{\aleph_0} \cdot 2^{\aleph_1}  $, which together with the previous direction gives us $\aleph_\omega^{\aleph_0} \cdot 2^{\aleph_1} = \aleph_\omega^{\aleph_1}$.
