TITLE: What is the product of all nonzero, finite cardinals?
QUESTION [6 upvotes]: To be specific, why does the following equality hold?
$$
\prod_{0\lt n\lt\omega}n=2^{\aleph_0}
$$

REPLY [5 votes]: 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.

REPLY [3 votes]: 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 $\displaystyle\prod_{n\in\omega}\omega$