For every infinite cardinal $\kappa$, $\kappa^2 =\kappa$. I can't seem to find a full proof that for every infinite cardinal $κ$, $k^2 = k$. Does anyone have one they could share?
 A: Claim 1. If $X$ is infinite and $Y$ is finite, $|X|+|Y|=|X|$.
Proof  Choose a countably infinite subset $A\subseteq X$.  Then there is a bijection $A\to A\sqcup Y$ (by shifting).  Combining this with the identity map $X-A\to X-A$ gives a bijection $X\to X\sqcup Y$.
Claim 2 If $X$ is infinite, $|X|+|X|=|X|$.
Proof Consider the collection of all bijections $A\sqcup A\to A$ where $A$ can be any subset of $X$, ordered by extension. They exist, for $X$ contains a countably infinite subset. Now use Zorn's lemma to pick a maximal element. Then $X-A$ must be finite, else we could glue up a bijection $Y\sqcup Y\to Y$ for $Y$ a countably infinite subset of $X-A$. The claim follows since then $A$ and $X$ are equinumerous by the first claim. 
In particular, by Schroder-Bernstein it follows that as long as at least one of $X$ and $Y$ is infinite we have $|X|+|Y|=\max(|X|,|Y|)$.
Claim 3 If $X$ is infinite, $|X\times X|=|X|$.
Proof. Again consider bijections $A\times A\to A$, and pick a maximal one. If $|X-A|\leq |A|$, then $|X|=\max(|A|,|X-A|)=|A|$, so choosing a bijection between $A$ and $X$ we're done. Hence we can assume $|X-A|\geq |A|$. Then $X-A$ contains a subset equinumerous with $A$; call it $Y$. We can then extend the bijection $A\times A\to A$ to a bijection $(A\cup Y)\times (A\cup Y)\to A\cup Y$ (here we use that $|A\times Y\cup Y\times A\cup Y\times Y|=|A\times A|+|A\times A|+|A\times A|=|A\times A|=|Y\times Y|$, with the second equality coming from Claim 2).  This contradicts the maximality of $A$.
