The answer is no. The continuum cannot be $\aleph_\omega$, and this can be proved in ZF, that is, without using the axiom of choice. To see this, suppose towards contradiction that $P(\omega)$ is equinumerous with $\aleph_\omega$. Since $P(\omega)^\omega$ P(\omega)$ is equinumerous with $P(\omega)$, P(\omega)^\omega$, and this does not require AC, it follows that there is a bijection $f:\aleph_\omega\cong (\aleph_\omega)^\omega$. Let $g(n)$ be the first ordinal $\alpha\lt\aleph_\omega$ that is not among $f(\beta)(n)$ for any $\beta\lt\aleph_n$. Since there are fewer than $\aleph_\omega$ many such $\beta$, it follows that there are fewer than $\aleph_\omega$ many such $f(\beta)(n)$, and so such an $\alpha$ exists. Thus, $g:\omega\to \aleph_\omega$. But notice that for any particular $\alpha\lt\aleph_\omega$, we have $\alpha\lt\aleph_n$ for some $n$ and consequently $g(n)=f(\alpha)(n)$, g(n)\neq f(\alpha)(n)$, and thus $g\neq f(\alpha)$. Thus, $f$ was not surjective to $(\aleph_\omega)^\omega$ after all, a contradiction.
This is just a standard proof of Konig's theorem (that $\aleph_\omega^\omega\gt\aleph_\omega$), and the point is that it doesn't use AC.
