Is the Axiom of Choice equivalent to say that for a (infinite) cardinal $\kappa$ it results $\kappa^2=\kappa$? 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 $\aleph\not\le\kappa$.
Lemma 10.5
If $\kappa$ is an infinite cardinal and $\aleph$ is an aleph, and if 
$$\kappa+\aleph=\kappa*\aleph,\tag{11.8}$$
then either $\kappa\ge\aleph$ or $\kappa\le\aleph$. In particular, if
$$\kappa+\aleph(\kappa)=\kappa*\aleph(\kappa)\tag{11.9}$$
then $\kappa$ is an aleph. 
Theorem 11.7
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 
$$\kappa+\aleph(\kappa)=\kappa*\aleph(\kappa).$$
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:
$$\kappa+\aleph(\kappa)=(\kappa+\aleph(\kappa))^2=\kappa^2+2\kappa*\aleph(\kappa) +(\aleph(\kappa))^2\ge\kappa*\aleph(\kappa).$$

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 $\kappa+\aleph(\kappa)\le\kappa*\aleph(\kappa)$.
 A: The strategy for proving 11.7 is to show that if $\kappa^2=\kappa$ for every infinite cardinal $\kappa$, then any infinite cardinal is an aleph, since this implies that every set can be well ordered, which is equivalent to $\mathsf{AC}$.
In particular, by 11.6, to show that every infinite cardinal $\kappa$ is an aleph it suffices to show that $\kappa+\aleph(\kappa)=\kappa\ast\aleph(\kappa)$ for every $\kappa$.
The inequality $\kappa+\aleph(\kappa)\leq\kappa\ast\aleph(\kappa)$ is clear from the definitions: $\kappa+\aleph(\kappa)=|(\kappa\times\{0\})\cup(\aleph(\kappa)\times\{1\})|$, while $\kappa\ast\aleph(\kappa)=|\kappa\times\aleph(\kappa)$| and so there is a clear injection from the former to the latter, given by $(\gamma,0)\mapsto(\gamma,a)$ and $(\eta,1)\mapsto(\eta,b)$, where $a,b$ are arbitrary distinct elements of $\kappa$ and $\aleph(\kappa)$ respectively.
The chain of inequalies on the last line of the quoted text shows that $\kappa+\aleph(\kappa)\geq\kappa\ast\aleph(\kappa)$, which together with the previous inequality establishes $\kappa+\aleph(\kappa)=\kappa\ast\aleph(\kappa)$, and this is enough to obtain $\mathsf{AC}$ by the proof strategy I outlined at the beginning of this answer.
