How to understand $\operatorname{cf}(2^{\aleph_0}) > \aleph_0$ As a corollary of  König's theorem, we have $\operatorname{cf}(2^{\aleph_0}) > \aleph_0$
.
On the other hand, we have $\operatorname{cf}(\aleph_\omega) = \aleph_0$.
Why the logic in the latter equation can't apply to the former one?
To be precise, why we can't have $\sup ({2^n:n<\omega}) = 2^{\aleph_0}$? 
 A: Should there be any reason to expect the cofinality to be a strictly increasing function? Perhaps naively, I have to admit that I don't remember my first thoughts on it right now.
Koenig's theorem assures us indeed that the continuum does not have cofinality $\omega$, from which we can deduce that $\aleph_\omega$ is never the cardinality of the continuum.
As for the second question, there is no reason to expect the continuum function to be continuous either (except the obvious similarity between the words). Indeed $2^n$ is finite, and so the supremum of all finite cardinals is $\aleph_0$ and not $2^{\aleph_0}$.
A: Note that in ordinal arithmetic $2^\omega$ is the supremum of $2^n$ for all $n<\omega$, so it is $\omega$ and thus has cofinality $\aleph_0$.
$\omega$ and $\aleph_0$ are the same set but they nevertheless behave differently in practice -- because tradition is to use the notation $\omega$ for operations where a limiting process is involved and $\aleph_0$ for operations where the "countably infinite" is used all at once in a single step. Cardinal exponentatiation $2^{\aleph_0}$ is such an all-at-once process.
A: We have $2^n < \omega$ for all $n < \omega$, so the supremum of the $2^n$ is the supremum of a set of ordinals cofinal in $\omega$, and hence is just $\omega$.
