TITLE: Why is ${\aleph_\omega}^{\aleph_1} = {\aleph_\omega}^{\aleph_0} \cdot {2}^{\aleph_1}$?
QUESTION [6 upvotes]: 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!

REPLY [4 votes]: 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 
\[ 
\aleph_\omega^{\aleph_1} = (2^{\aleph_1} \cdot \aleph_\omega)^{\aleph_0} = 2^{\aleph_1} \cdot \aleph_\omega^{\aleph_0}. \]