What is $\aleph_0!$? What is $\aleph_0!$ ? 
I know that in the original definition the factorial is defined for natural numbers but, what if we extend this concept to cardinal numbers? 
This concept has been extended to the real numbers by the $\Gamma$ function but I never see this kind of extension before. 
This is a proof that I made by myself and can be incorrect but still interesting for me.
$\aleph_0\times(\aleph_0 - 1)\times(\aleph_0 - 2)\times ...$
We can rewrite this as
$$\aleph_0! = \prod_{i = 1}^{\infty}(\aleph_0 - i) = \prod_{i = 1}^{\infty}(\aleph_0)$$ 
But, is this equal to:
$$\aleph_0^{\aleph_0}$$ 
Also, if we assume the continumm hypothesis
$2^{\aleph_0} = \mathfrak{c} \leq \aleph_0^{\aleph_0} \leq \mathfrak{c}$
Hence, $\aleph_0! = \mathfrak{c}$

We have $k!=1\cdot2\cdots k$, i.e., it is products of all numbers with size at most $k$.
Therefore 
$$\aleph_0! = 1\cdots 2 \cdots \aleph_0 = \prod_{k\le\aleph_0} k$$
seems like a possible generalization. (Although probably taking the number of bijections - as suggested in other answers - is a more natural generalization.)
I will add that the above product has two - it can be reformulated in this way: For each $k\le\aleph_0$ we have a set $A_k$ such that $|A_k|=k$. And we are interested in the cardinality of the Cartesian product of these sets $\prod_{k\le\alpha_0} A_k$.
It is not difficult to see that if we use this definition, then
$$2^{\aleph_0} \le \aleph_0! \le \aleph_0^{\aleph_0}.$$
Together with $\aleph_0^{\aleph_0}=2^{\aleph_0}$, we get $\aleph_0! = 2^{\aleph_0}$.

If we wanted to do similar generalization for higher cardinalities, we could define
$$\kappa! = \prod_{\alpha\le\kappa} |\alpha|$$
where the product is taken over all ordinals $\le\kappa$. Then exactly the same argument shows that $2^\kappa \le \kappa! \le \kappa^\kappa$.
