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.
Theorem I: If $X$ is a set, then $X$ can be well-ordered if and only if there exists a choice function on $\mathcal P(X)\setminus\{\varnothing\}$.
Proof. If $X$ can be well-ordered fix a well-ordering, and a choice function returns the minimal element of every non-empty subset.
If $\mathcal P(X)\setminus\{\varnothing\}$ has a choice function $F$ we define by a transfinite argument a well-ordering of $X$:
Suppose for $\alpha$, $x_\beta$ was chosen for $\beta<\alpha$, let $x_\alpha=F(X\setminus\{x_\beta\mid\beta<\alpha\})$. This is well-defined, and obviously one-to-one from ordinals into $X$. Since $X$ is a set, the induction has to end at some ordinal $\kappa$. Therefore we have a bijection between $\kappa$ and $X$ so $X$ can be well-ordered. $\square$
Theorem II: It is consistent with ZF that the real numbers cannot be well-ordered.
I won't prove that here, since this requires a lot more machinery from advanced set theory, but this is essentially what Cohen proved in his original work about forcing.
Corollary: It is consistent with ZF that there is no choice function on $\mathcal P(\mathbb R)\setminus\{\varnothing\}$.
This is now a trivial corollary, in a model where $\mathbb R$ cannot be well-ordered -- there is no choice function on $\mathcal P(\mathbb R)\setminus\{\varnothing\}$.
Best Answer
Yes because there is one $\aleph_\alpha\to \aleph_\alpha\times\aleph_\alpha$ without the axiom of choice : the proof of this fact does not use it.
Moreover the proof of "if there is a bijection $X\to Y$ then there is one $\mathcal{P}(Y)\to\mathcal{P}(X)$" does not use the axiom of choice either.