Following what is written at the 11th chapter of "The Axiom of Choice" by Thomas Jech.
For every infinite cardinal numer $\kappa$, let $\aleph(\kappa)$ be the Hartogs number of $\kappa$, i.e., the least ordinal which cannot be embedded by a one-to-one mapping in a set of cardinality $\kappa$. For every $\kappa$, $\aleph(\kappa)$ is an aleph, viz. the least aleph $\aleph$ such that $\aleph\not\le\kappa$.
If $\kappa$ is an infinite cardinal and $\aleph$ is an aleph, and if
then either $\kappa\ge\aleph$ or $\kappa\le\aleph$. In particular, if
then $\kappa$ is an aleph.
If $\kappa^2=\kappa$ for every infinite cardinal number $\kappa$, then the Axiom of Choice holds.
Proof. We will show that under the assumption of the theorem, every infinite cardinal is an aleph. To do so, it suffices to show that
Since $\kappa+\aleph(\kappa)\le\kappa*\aleph(\kappa)$, we have only to show that $\kappa+\aleph(\kappa)\ge\kappa*\aleph(\kappa)$.
This is proved as follows:
So I don't understand the proof of theorem 11.7: could someone explain to me why what is here written prove the theorem?
I have understood that if every infinite cardinal number is an aleph also every infinite set can put in bijection with an aleph, which is a well ordered set so this implies the well-ordering theorem and this is equivalent to the Choice Axiom; but how to prove the opposite implication?
Then I don't understand why $\kappa+\aleph(\kappa)\le\kappa*\aleph(\kappa)$.