TITLE: What is the product of finitely indexed alephs?
QUESTION [6 upvotes]: I'm simply curious about why the following equality holds:
$
\displaystyle\prod_{n\lt\omega}\aleph_n=\aleph_\omega^{\aleph_0}.
$
Much thanks!

REPLY [6 votes]: The following is a theorem of Tarski: $$\prod_{\alpha<\kappa}\lambda_\alpha = \bigg(\sup\{\lambda_\alpha\mid\alpha<\kappa\}\bigg)^\kappa$$
From which it follows trivially that $\prod_\omega\aleph_n = \aleph_\omega^{\aleph_0}$. Simply let $\kappa=\omega$ and $\lambda_n=\aleph_n$, since $\sup\{\aleph_n\mid n<\omega\}=\aleph_\omega$ we have that the the product of the $\aleph_n$'s equals to $\aleph_\omega^{|\omega|}=\aleph_\omega^{\aleph_0}$.
The proof of the theorem can be found in details in various books (currently I am going through Introduction to Cardinal Arithmetics, it appears as Lemma 1.6.15).
The proof in a nutshell:
First let $\mu=\sup\{\lambda_\alpha\mid\alpha<\kappa\}$. The idea behind the proof is to consider a partition of $\kappa$ into $\{A_\xi\mid\xi<\kappa\}$ (simply by the fact that $\kappa\cdot\kappa = \kappa$), then consider the product over all the parts, that is: $$\mu^\kappa = \prod_{\alpha<\kappa}\mu \le\prod_{\xi<\kappa}\ \ \prod_{\alpha\in A_\xi}\lambda_\alpha=\prod_{\alpha<\kappa}\lambda_\alpha\le\prod_{\alpha<\kappa}\mu=\mu^\kappa$$
The first $\le$ sign comes from the distributivity of products over such partitions of the index set, and the fact that for every $A_\xi$ we have that $\sup A_\xi=\mu$.