Let me begin with a remark related to the question of whether . We showed that this is the case if for some , or if is Dedekind-finite.
Theorem. The axiom of choice is equivalent to the statement that any Dedekind-infinite cardinal is a square.
Proof. Let be a set. Assuming that every D-infinite cardinal is a square, we need to show that is well-orderable. We may assume that . Otherwise, replace with . Let . Assume that is a square, say . Then . By Homework problem 2, , so for some , and .
Lemma. Suppose are D-infinite sets and is an (infinite) initial ordinal. If then either or .
Proof. Let be an injection. If there is some such that we are done, so we may assume that for all there is some such that . Letting be the least such , the map is an injection of into .
By the lemma, it must be that either or else . The former is impossible since , so is well-orderable, and thus so is , and since , then is well-orderable as well.