# Is there an easy proof for ${\aleph_\omega} ^ {\aleph_1} = {2}^{\aleph_1}\cdot{\aleph_\omega}^{\aleph_0}$?

The question contains 2 stages:

1. Prove that ${\aleph_n} ^ {\aleph_1} = {2}^{\aleph_1}\cdot\aleph_n$

This one is pretty clear by induction and by applying Hausdorff's formula.

1. Prove ${\aleph_\omega} ^ {\aleph_1} = {2}^{\aleph_1}\cdot{\aleph_\omega}^{\aleph_0}$

Is there an easy proof for the second one?

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As you mention, the first equation is a consequence of Hausdorff's formula and induction.

For the second: Clearly the right hand side is at most the left hand side. Now: Either $2^{\aleph_1}\ge\aleph_\omega$, in which case in fact $2^{\aleph_1}\ge{\aleph_\omega}^{\aleph_1}$, and we are done, or $2^{\aleph_1}<\aleph_\omega$.

I claim that in this case we have ${\aleph_\omega}^{\aleph_1}={\aleph_\omega}^{\aleph_0}$. Once we prove this, we are done.

Note that $\aleph_\omega={\rm sup}_n\aleph_n\le\prod_n\aleph_n$, so $${\aleph_\omega}^{\aleph_1}\le\left(\prod_n\aleph_n\right)^{\aleph_1}=\prod_n{\aleph_n}^{\aleph_1}.$$

Now use part 1, to conclude that ${\aleph_n}^{\aleph_1}<\aleph_\omega$ for all $n$, since we are assuming that $2^{\aleph_1}<\aleph_\omega$. This shows that the last product is at most $\prod_n \aleph_\omega={\aleph_\omega}^{\aleph_0}$.

This shows that ${\aleph_\omega}^{\aleph_1}\le {\aleph_\omega}^{\aleph_0}$. But the other inequality is obvious, and we are done.

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