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).

Answer 1

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.