You've made a very big mess, I think. The way you defined cardinal numbers is very awkward, so to say. In my eyes, anyway.
Let us review the construction of ordinals:
- $0 = \emptyset$
- $\alpha+1 = \alpha \cup \{\alpha\}$
- At limit stages, $\delta = \bigcup_{\beta<\delta} \beta$
Now we define initial ordinals as ordinal numbers which cannot be bijected with any smaller ordinal. For example $\omega$ (the set of natural numbers) is such ordinal, while $\omega+1, \omega+\omega, \omega\cdot\omega$ are not initial ordinals.
Under the axiom of choice, every set is well-orderable, and therefore we can choose the initial ordinal out of each equivalence class as a representative. This is the usual notion of $\aleph$ numbers under the axiom of choice.
Without assuming choice, the cardinal system is not well-ordered and can behave very strangely.
Regardless to that, when you are only dealing with ordinals you don't need choice because there is a canonical choice function (take the minimal element). So even without the axiom of choice it is true that $\omega_\alpha$ has no bijection with $\omega_\beta$ for $\alpha\not=\beta$.
The idea behind aleph numbers, as far as I see it, is that it is a well ordering of cardinalities (not necessarily all cardinalities, though) and as such it holds just fine even when not assuming choice. However, in the case you don't have the axiom of choice to help you out, $2^{\aleph_0}$ might not be well-orderable and thus won't be represented by an ordinal, and therefore won't be represented by an $\aleph$ number, same with multiplication. It is equivalent to the axiom of choice that for every infinite set $|X| = |X\times X|$.
Just last remark, you said that the construction you gave infers the existence of $\aleph_n$ for every natural number $n$ while in fact it gives you $\aleph_\alpha$ for every ordinal $\alpha$ and not just for the natural numbers.
I don't know any interesting consequences of $2^{\aleph_0}=\aleph_2$ or related statements alone. However, there are reasonably natural principles which imply $2^{\aleph_0}=\aleph_2$ and also imply many interesting statements, my favorite being the Proper Forcing Axiom (PFA) and its variants; see this paper by Moore.
Meanwhile, the principle "For all $\alpha$, $2^{\aleph_\alpha}=\aleph_{\alpha+2}$" is consistent (relative to large cardinals, necessarily), but that's far from obvious; see this paper of Magidor and Woodin, as well as this Mathoverflow question.
That said, again I don't know any interesting consequences from this alone. In general, my understanding is that simple arithmetic facts like these don't tend to have lots of interesting consequences on their own. (Obviously when I say "simple" I'm referring to the form of the statement; as the results mentioned above indicate, "$\forall\alpha(2^{\aleph_\alpha}=\aleph_{\alpha+2})$" is extremely complicated once we really dive intothings.)
Best Answer
Mike: If you fix an ordinal $\alpha$, then it is consistent that ${\mathfrak c}>\aleph_\alpha$. More precisely, there is a (forcing) extension of the universe of sets with the same cardinals where the inequality holds.
If you begin with a model of GCH, then you can go to an extension where ${\mathfrak c}=\aleph_\alpha$ and no cardinals are changed, as long as $\alpha$ is not a limit ordinal of countable cofinality. For example, $\aleph_{\aleph_\omega}$ is not a valid size for the continuum. But it can be larger.
Here, the cofinality of the limit ordinal $\alpha$ is the smallest $\beta$ such that there is an unbounded function $f:\beta\to\alpha$. There is a result of König that says that $\kappa^\lambda>\kappa$ if $\lambda$ is the cofinality of $\kappa$. If $\kappa={\mathfrak c}$, this says that $\lambda>\omega=\aleph_0$, since ${\mathfrak c}=2^{\aleph_0}$ and $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$. Since $\aleph_{\aleph_\omega}$ has cofinality $\omega$, it cannot be ${\mathfrak c}$.
But this is the only restriction! The technique to prove this (forcing) was invented by Paul Cohen and literally transformed the field.