Proving cardinal inequality

Question

How does one show that $\prod_{n < \omega}\aleph_n \leq \aleph^{\aleph_0}_{\omega}$?

Answer 1

If we think of cardinals as sets then there is inclusion. A general element in $\prod_{n<\omega}\aleph_n$ is a function $f:\omega\to \cup_{n<\omega} \aleph_{n}$ which satisfies $f(n)\in\aleph_n$ for each $n<\omega$. But such $f$ is also a function from $\omega$ to $\aleph_{\omega}$ (because $\aleph_n\subseteq\aleph_{\omega}$ for each $n$) and hence belongs to the set of functions from $\aleph_0$ to $\aleph_{\omega}$.