Is aleph-omega a Mahlo cardinal? So I've read up a little on Mahlo cardinals and found a definition along the lines of of $\kappa$ is Mahlo iff it is inaccessible and regular cardinals below it form a stationary set.
Using this definition, I can see why $\aleph_0$ and $\aleph_1$ are not Mahlo, in that the first is not inaccessible because it is countable and that the second is a successor cardinal (and therefore not inaccessible).
But then I get to $\aleph_\omega$... Which is inaccessible, because it is uncountable and a weak limit cardinal. The stationary subset part is tripping me up though. The cardinals below it are $\{\aleph_0, \aleph_1, ..., \aleph_n, ... | n < \omega\}$.
So I guess, is $\aleph_\omega$ Mahlo? If yes, why? And if no, why not? Further, what is the "smallest" Mahlo cardinal?

A Mahlo cardinal has to be regular, which $\aleph_\omega$ is not.  $\aleph_\omega =\bigcup\aleph_n$, so $\operatorname{cf}(\aleph_\omega)=\aleph_0$.  Every strong inaccessible $\kappa$ satisfies $\kappa=\aleph_\kappa$, but even that is not enough as the lowest $\kappa$ satisfying that has $\operatorname{cf}(\kappa)=\aleph_0$.  As we can't prove even that strong inaccessibles exist, we can't say where they are in the $\aleph$ heirarchy.
