A cardinality number to the square is itself?

Question

Is there a cardinal number $\kappa$ so that $\kappa \cdot \kappa$ (which is the cardinality of the set of the Cartesian product of $\kappa$ by itself) is not equivalent to $\kappa$?

My progress: I remember learning about $Q \cdot Q$ to be $Q$, and we can prove it by the snailing technique (Q is the rationals). Then I proved the same for $R$, the real numbers (I assumed continuum cardinality).

Of course , you mean infinite cardinals. The claim would be false for finte cardinals. Your claim should be true, but I do not know how to rigorously prove it. — Peter, Aug 31 '18 at 22:42
It is true for all infinite cardinals. There is a standard proof by transfinite induction. You can find most books on set theory or here http://planetmath.org/idempotencyofinfinitecardinals — spaceisdarkgreen, Aug 31 '18 at 22:55
It is true for all infinite well-orderable cardinals, which would be all infinite cardinals if the Axiom of Choice holds. In fact, $\kappa\cdot\kappa=\kappa$ holds for all infinite cardinals $\kappa$ if and only if the Axiom of Choice holds, as you can see here. — Cameron Buie, Aug 31 '18 at 23:15

score 4 · Answer 1 · edited Sep 01 '18 at 00:38

The question has been answered by Cameron Buie's comment. I wrote this answer to make it visible at first glance that the question is no longer open.

Cameron's comment (paraphrased):

$A\times A \simeq A$ holds for all well-orderable infinite sets $A$. If the axiom of choice holds, every set is well-orderable, so this means the result holds for all infinite sets. However, this is not provable in the absence of choice, and in fact $A\times A\simeq A$ holds for any infinite set $A$ if and only if the axiom of choice holds, as you can see here.

A cardinality number to the square is itself?

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