What is $\aleph_0!$?

First, a couple quick comments:

• The continuum hypothesis isn't needed (and I'm not sure how you used it) - $(\aleph_0)^{\aleph_0}=2^{\aleph_0}$, provably in ZF (we don't even need the axiom of choice!).

• Also, subtraction isn't really an appropriate operation on cardinals - while it's clear what $\kappa-\lambda$ should be if $\lambda<\kappa$ and $\kappa$ is infinite, what is $\aleph_0-\aleph_0$?

The right definition of the factorial is as the size of the corresponding group of permutations: remember that in the finite case, $n!$ is the number of permutations of an $n$-element set, and this generalizes immediately to the $\kappa$-case. It's now not hard to show that $\kappa!=\kappa^\kappa$ in ZFC - that is, there is a bijection between the set of permutations of $\kappa$ and the set of all functions from $\kappa$ to $\kappa$.

And this can be simplified further: it turns out $\kappa^\kappa=2^\kappa$, always. Clearly we have $2^\kappa\le\kappa^\kappa$, and in the other direction $$\kappa^\kappa\le (2^\kappa)^\kappa=2^{\kappa\cdot\kappa}=2^\kappa.$$

By the way, the equality $\kappa^\kappa=2^\kappa$ can be proved in ZF alone as long as $\kappa$ is well-orderable, the key point being that Cantor-Bernstein doesn't need choice.

You can define the factorial of a cardinal $c$ to be the cardinality of the set of bijections $X \to X$ where $X$ is a set with cardinality $c$; this reproduces the usual factorial if $X$ is finite.

So $\aleph_0!$ is the cardinality of the set of bijections $\mathbb{N} \to \mathbb{N}$, and it's not hard to show that this has cardinality $\aleph_0^{\aleph_0}$, for example using Cantor-Bernstein-Schroder.