What is the product of finitely indexed alephs? I'm simply curious about why the following equality holds:
$
\displaystyle\prod_{n\lt\omega}\aleph_n=\aleph_\omega^{\aleph_0}.
$
Much thanks!
 A: 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 $\sup A_\xi=\mu$.
A: It is obvious that $\prod_{n<\omega}\aleph_n \preceq (\aleph_\omega)^\omega$, since the LHS is a subset of the RHS.
For the other direction, let's try to construct a injection from $(\aleph_\omega)^\omega$ (that is, the set of infinite sequences of elements of $\aleph_\omega$) into $\prod_{n<\omega}\aleph_n$. Without loss of generality we can assume that $0$ does not occur in the original sequence (say, by adding $1$ to every finite element). Now $\aleph_\omega = \bigcup_{n<\omega} \aleph_n$, so for each $\alpha\in\aleph_\omega$ there is a least $N<\omega$ such that $\alpha\in\aleph_n$ for all $n\geq N$. Thus, for each $0$-free infinite sequence of elements of $\aleph_\omega$, we can pad that sequence with $0$'s such that each element comes late enough to be in $\aleph_n$ for $n$ being its index. The nonzero elements of the padded sequence are exactly the elements of the original sequence, and in the same order. So we have an injection from $(\aleph_\omega)^\omega$ to $\prod_{n<\omega}\aleph_\omega$.
Now apply Cantor-Bernstein.
