# Why is ${\aleph_\omega}^{\aleph_1} = {\aleph_\omega}^{\aleph_0} \cdot {2}^{\aleph_1}$? [duplicate]

I am supposed to prove that ${\aleph_\omega}^{\aleph_1} = {\aleph_\omega}^{\aleph_0} \cdot {2}^{\aleph_1}$ , but I really have no idea how to start or what to do. I thought I could use the following fact: ${2}^{\aleph_1}= {\aleph_1}^{\aleph_1}$, because of the infiniteness of ${\aleph_1}$.

I hope someone will show me how this works. Thanks in advance!

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## marked as duplicate by Lord_Farin, Calvin Lin, Amzoti, TMM, O.L.May 29 '13 at 15:32

We have, since $\aleph_\omega$ is a limit, that $\aleph_\omega^{\aleph_1} = (\sup_n \aleph_n^{\aleph_1})^{\operatorname{cf}\aleph_\omega}$ By Hausdorff and induction $\aleph_{n+1}^{\aleph_1} = \aleph_n^{\aleph_1} \cdot \aleph_{n+1} = \aleph_1^{\aleph_1}\cdot \aleph_{n+1} = 2^{\aleph_1} \cdot \aleph_{n+1}$ Hence $\sup_n \aleph_n^{\aleph_1} = 2^{\aleph_1} \cdot \aleph_{\omega}$ As $\operatorname{cf}\aleph_\omega = \aleph_0$, finally $\aleph_\omega^{\aleph_1} = (2^{\aleph_1} \cdot \aleph_\omega)^{\aleph_0} = 2^{\aleph_1} \cdot \aleph_\omega^{\aleph_0}.$