How do I go about showing:
$\aleph_{\omega}^{\aleph_{1}}$=$\aleph_{\omega}^{\aleph_0}$.$2^{\aleph_1}$
My first thought was to apply Hausdorff's formula directly, but that doesn't quite work. The book provides a hint using products and sums that I don't quite understand. Can you please help?
The question is from Introduction to Set Theory by Hrbacek and Jech.
That we have a product on the right-hand side is an indication that there are two cases to be distinguished. Note that it will clearly suffice to prove $\aleph_\omega^{\aleph_1} \le \aleph_\omega^{\aleph_0}2^{\aleph_1}$.
Suppose first that for some $\kappa < \aleph_\omega$, we have $\kappa^{\aleph_1} \ge \aleph_\omega$. Then $\kappa^{\aleph_1} \le \aleph_\omega^{\aleph_1} \le (\kappa^{\aleph_1})^{\aleph_1} = \kappa^{\aleph_1}$. I leave it to you to use Hausdorff's formula to establish $\kappa^{\aleph_1} = 2^{\aleph_1}$.
Now suppose that $\kappa^{\aleph_1} < \aleph_\omega$ for all $\kappa <\aleph_\omega$. Then since $\aleph_1 \ge \operatorname{cf}\, \aleph_\omega = \aleph_0$, it is known that $\aleph_\omega^{\aleph_1} = \aleph_\omega^{\operatorname{cf}\aleph_\omega} = \aleph_\omega^{\aleph_0}$.
This latter fact appears as Theorem 5.20(iii)(b) in Jech's Set Theory. I can add a proof of this if you want, but I guess it's in Hrbacek-Jech as well.
