= Square of an infinite cardinal
Assuming the <axiom of choice>, every infinite cardinal satisfies
$$
\kappa\cdot\kappa=\kappa.
$$
One proof takes a least counterexample $\kappa$, identifies cardinals with <initial ordinal>[initial ordinals], and well-orders $\kappa\times\kappa$ first by $\max\{\alpha,\beta\}$. Every proper initial segment then has cardinality below $\kappa$ by minimality, so the whole order has cardinality at most $\kappa$, contradicting the choice of $\kappa$.
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