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Under the axiom of choice, identify every infinite cardinal $\kappa$ with its initial ordinal. We prove ${\kappa }^2=\kappa$ by transfinite induction on infinite cardinals. The countable case follows from the usual diagonal enumeration of $\mathbb{N}\times \mathbb{N}$.

For the induction step, well-order the Cartesian product $\kappa \times \kappa$ by first comparingand then using lexicographic order among pairs with the same maximum. The predecessors of $(\alpha ,\beta )$ lie in $(\eta +1)\times (\eta +1)$, where $\eta =\max \{\alpha ,\beta \}<\kappa$. Put $\mu =|\eta +1|<\kappa$. If $\mu$ is infinite, the induction hypothesis makes this predecessor set have cardinal at most ${\mu }^2=\mu <\kappa$; if $\mu$ is finite, the same conclusion is immediate.

Let $\theta$ be the order type of this well-order. Every proper initial segment of $\theta$ therefore has cardinal less than $\kappa$. Hence $\theta <{\kappa }^{+}$, since otherwise its initial segment of order type $\kappa$ would have cardinal $\kappa$. Thus $|\kappa \times \kappa |=|\theta |\le \kappa$. The map $\alpha \mapsto (\alpha ,0)$ gives the reverse injection, and the Cantor-Schröder-Bernstein theorem yields

This is König theorem for cardinal numbers. Since ${\kappa }_i<{\lambda }_i$, map $(i,\alpha )$ in the disjoint union to the element $s_{i,\alpha }$ of the product defined byHere $\alpha <{\kappa }_i$ implies $\alpha +1<{\lambda }_i$. The unique nonzero coordinate and its value recover $(i,\alpha )$, so this is an injection

For strictness, suppose $F$ mapped the disjoint union onto the product. For each $i$, the sethas cardinality at most ${\kappa }_i$, so it cannot exhaust ${\lambda }_i$. Choose $b_i\in {\lambda }_i\setminus A_i$. Then $b=(b_i{)}_{i\in I}$ differs from $F(i,\alpha )$ in coordinate $i$ for every $(i,\alpha )$, contradicting surjectivity. By Cantor-Schröder-Bernstein theorem, an injection in the reverse direction would combine with the displayed injection to give a bijection and hence a surjection. Therefore