Show $\sup_\limits{n\in\omega}\prod_\limits{i\in n}2=\aleph_{0}$

Question

I would appreciate help showing:

$$\sup_{n\in\omega} \prod_{i\in n} 2$$

I think $\prod_\limits{i\in n}2=2^n$. If this is correct, then I am stuck on what $\sup_\limits{n\in\omega}2^{n}$ is.

To the extent that $\omega$ is the $\sup$ of the natural numbers, then it would be $2^\omega$ or $2^{\aleph_0}$. But I don't think this is correct.

Thanks

Answer 1

$$\sup_{n\in\omega} 2^n = \omega$$

It is obvious that the supremum could not be any finite number (there is always a power of two greater than it).

$\omega$ is greater than every $2^n$, and is the smallest infinite ordinal.

Therefore, it is the supremum.