What is the product of all nonzero, finite cardinals?

Question

To be specific, why does the following equality hold? $$ \prod_{0\lt n\lt\omega}n=2^{\aleph_0} $$

Answer 1

As a product of cardinals, yes:

$$2^{\aleph_0} \leq \prod_{0 < n < \omega} n \leq {\aleph_0}^{\aleph_0} \leq 2^{\aleph_0 \cdot \aleph_0} = 2^{\aleph_0}$$

As a product of ordinals, no:

$$\prod_{0 < n < \omega} n \leq \prod_{0 < n < \omega} \omega = {\omega}^{\omega}$$ but the ordinal ${\omega}^{\omega}$ is countable.

Answer 2

If $\displaystyle f\in\prod_{n\in\omega} 2$ then $f(n)\in\{0,1\}$, and in particular for $n>1$ we have that $f(n)\in n$. Therefore this is a proper subset of $\displaystyle f\in\prod_{0<n<\omega} n$, therefore the cardinality is at least continuum.

On the other hand $\omega^\omega$ has cardinality continuum, and the same argument shows that the product is a subset of $\displaystyle\prod_{n\in\omega}\omega$