The proof shall proceed by Transfinite Induction on $x$. Let: :$\forall y \in x: y < \omega \lor \left|{ y }\right| = \left|{ y \times y }\right|$ There are two cases: === Case 1: $\left|{ y }\right| = \left|{ x }\right|$ for some $y \in x$ === If this is so, then: {{begin-eqn}} {{eqn | l = \left\vert{ x }\right\vert       | r = \left\vert{ y }\right\vert       | c =  }} {{eqn | r = \left\vert{ y \times y }\right\vert       | c = Inductive hypothesis }} {{eqn | r = \left\vert{ x \times x }\right\vert       | c =  }} {{end-eqn}} {{qed|lemma}} === Case 2: $\left|{ y }\right| < \left|{ x }\right|$ for all $y \in x$ === We have that either: :$y < \omega$ or: :$\left|{ y }\right| = \left|{ y + 1 }\right|$ In either case, we have that: :$\left|{y + 1}\right < \left|{x}\right|$ and therefore: :$y + 1 \in x$ Therefore, $x$ is a limit ordinal. Let $R_0$ denote the canonical order of $\operatorname{On}^2$. Let $J_0$ be defined as the unique order isomorphism between $\operatorname{On}^2$ and $\operatorname{On}$ as defined in canonical order. {{refactor|The above section Definition:Canonical Order#Properties no longer exists. Instead, this link is to go to the (not yet established) second definition of Canonical Order.}} It follows that: :$J_0 \left({x \times x}\right) = \bigcup_{y \mathop \in x} \bigcup_{z \mathop \in x} J_0 \left({y, z}\right)$ But moreover: {{begin-eqn}} {{eqn | l = J_0 \left({ y,z }\right)       | r = \in^{-1} \left({ J_0 \left({ y,z }\right)}\right)       | c = Definition of Preimage }} {{eqn | r = J_0 \left({ R_0^{-1} \left({ y,z }\right) }\right)       | c = as $J_0$ forms an Order Isomorphism. }} {{end-eqn}} Since $J_0$ is a bijection: {{begin-eqn}} {{eqn | l = \left\vert{J_0 \left({y, z}\right) }\right\vert       | r = \left\vert{J_0 \left({R_0^{-1} \left({y, z}\right) }\right) }\right\vert       | c = Equality shown above }} {{eqn | r = \left\vert{ R_0^{-1} \left({y, z}\right) }\right\vert       | c = Image of Bijection Cardinal Equality }} {{end-eqn}} Take $\max \left({y, z}\right)$. {{begin-eqn}} {{eqn | l = \left({a, b}\right)       | o = \in       | r = R_0^{-1} \left({y, z}\right)       | c = hypothesis }} {{eqn | ll= \iff       | l = \left({a, b}\right)       | o = R_0       | r = \left({y, z}\right)       | c = Definition of Image }} {{eqn | ll= \iff       | l = a, b       | o = \le       | r = \max \left({y, z}\right)       | c = Definition of Canonical Order }} {{eqn | ll= \iff       | l = a, b       | o = \in       | r = \max \left({y, z}\right) + 1       | c = Definition of Successor }} {{end-eqn}} It follows that: {{begin-eqn}} {{eqn | l = \left\vert{R_0^{-1} \left({y, z}\right) }\right\vert       | o = \le       | r = \left\vert{\max \left({y, z}\right) + 1 \times \max \left({y, z}\right) + 1}\right\vert        | c =  }} {{eqn | r = \left\vert{\max \left({y, z}\right) + 1}\right\vert       | c = Inductive hypothesis }} {{eqn | o = <       | r = \left\vert{x}\right\vert       | c = Hypothesis for the case }} {{end-eqn}} Therefore: :$\left\vert{ R_0^{-1} \left({y, z}\right) }\right\vert < \left|{ x }\right|$ Thus by Cardinal Inequality implies Ordinal Inequality: :$\forall y, z \in x: J_0 \left({y, z}\right) < \left|{x}\right|$ It follows by Supremum Inequality for Ordinals that: :$J_0 \left({x \times x}\right) \subseteq \left|{x}\right|$ Hence: {{begin-eqn}} {{eqn | l = \left\vert{x \times x}\right\vert       | r = \left\vert{J_0 \left({x \times x}\right) }\right\vert       | c = Image of Bijection Cardinal Equality }} {{eqn | o = \le       | r = \left\vert{x}\right\vert       | c = Subset of Ordinal implies Cardinal Inequality }} {{end-eqn}} But also by Set Less than Cardinal Product: :$\left|{x}\right| \le \left|{x \times x}\right|$ Thus: :$\left|{x}\right| = \left|{x \times x}\right|$ {{qed}}