What is the cardinality of $(\aleph_\omega)^{\aleph_0}$?

Question

I want to show that $\aleph_0 \cdot \aleph_1 \cdot \aleph_2 \cdots = (\aleph_\omega)^{\aleph_0}$. I am not sure why this is true. My thinking is that $\aleph_n \cdot \aleph_{n+1} = \max\{\aleph_n,\aleph_{n+1}\} = \aleph_{n+1}$ and so $\aleph_0 \cdot \aleph_1 \cdot \aleph_2 \cdots = \sup\{\aleph_n \mid n \in \mathbb{N}\} = \aleph_\omega$. Now we want to show that $\aleph_\omega = (\aleph_\omega)^{\aleph_0}$. I read in Jech that for cardinals $\kappa$ and $\lambda$ with $\lambda \geq cf(\kappa)$ we have $\kappa^\lambda > \kappa$. Since the cofinality of $\aleph_\omega$ is $\aleph_0$ then we conclude that $(\aleph_\omega)^{\aleph_0} > \aleph_\omega$ and so the result I am trying to prove does not hold. Clearly I have made a mistake somewhere but cannot figure out what it is. Any help is much appreciated.

Answer 1

Your mistake is assuming that infinite products are continuous, i.e. that the product is the supremum of its initial segments.

By that logic, $\aleph_0^{\aleph_0}$ is $\sup\{\aleph_0^n\mid n<\omega\}=\aleph_0$.

As to your question, what is the cardinality of $(\aleph_\omega)^{\aleph_0}$? Well, we only know that it is at most $2^{\aleph_0}\cdot\aleph_{\omega_4}$. But it could be that $2^{\aleph_0}<\aleph_\omega$ and $(\aleph_\omega)^{\aleph_0}$ is, assuming the consistency of large cardinal axioms, any $\aleph_{\alpha+1}$ where $\alpha$ is a countably infinite ordinal.