Ordinal exponentiation - $2^{\omega}=\omega$
Ordinal exponentiation is not cardinal exponentiation.
The cardinal exponentiation $2^\omega$ is indeed uncountable and has the cardinality of the continuum.
The ordinal exponentiation $2^\omega$ is the supremum of $\{2^n\mid n\in\omega\}$ which in turn is exactly $\omega$ again.
Also related:
- How is $\epsilon_0$ countable?
- Do $\omega^\omega=2^{\aleph_0}=\aleph_1$?
$2^\omega$ means two different things, depending on whether you’re doing cardinal exponentiation or ordinal exponentiation. You’re thinking of cardinal exponentiation: for that it’s perfectly true that $2^\omega>\omega$. For ordinal exponentiation, however, $2^\omega$ is simply $\bigcup_{n\in\omega}2^n=\omega$.
Note that $2^\omega$ is not the powerset of $\omega$, and $|2^\omega|\ne2^{|\omega|}$, where the alatter is cardinal arithmetic. If you read on in that Wikipedia article, you'll see that $2^\omega$ is obtained by taking the maps $\omega\to 2$ with finite support, not arbitrary such maps. That corresponds to the finite subsets rather than arbitrary subsets, thus to terminating binary sequences (esp. ratioanl numbers) instead of arbitrary binary sequences (real numbers) if you like.