The proof shall proceed by Transfinite Induction on $z$: === Basis for the Induction === Let $0$ denote the zero ordinal. {{begin-eqn}} {{eqn | l = x \times \left({y \times 0}\right)       | r = x \times 0       | c = Ordinal Multiplication by Zero }} {{eqn | r = 0       | c = Ordinal Multiplication by Zero }} {{eqn | r = \left({x \times y}\right) \times 0       | c = Ordinal Multiplication by Zero }} {{end-eqn}} This proves the basis for the induction. === Induction Step === {{begin-eqn}} {{eqn | l = x \times \left({y \times z}\right)       | r = \left({x \times y}\right) \times z       | c = Inductive Hypothesis }} {{eqn | ll = \implies       | l = x \times  \left({y \times z^+}\right)       | r = x \times  \left({\left({y \times z}\right) + y}\right)       | c = Definition of Ordinal Multiplication }} {{eqn | r = x \times \left({y \times z}\right) + \left({x \times y}\right)       | c = Ordinal Multiplication is Left Distributive }} {{eqn | r = \left({x \times y}\right) \times z + \left({x \times y}\right)       | c = Inductive Hypothesis }} {{eqn | r = \left({x \times y}\right) \times z^+       | c = Definition of Ordinal Multiplication }} {{end-eqn}} This proves the induction step. === Limit Case === The inductive hypothesis for the limit case states that: :$\forall w \in Z: x \times \left({y \times w}\right) = \left({x \times y}\right) \times w$ where $z$ is a limit ordinal. The proof shall proceed by cases: === Case 1 === If $y = 0$, then: {{begin-eqn}} {{eqn | l = x \times \left({y \times z}\right)       | r = x \times 0       | c = Ordinal Multiplication by Zero }} {{eqn | r = 0       | c = Ordinal Multiplication by Zero }} {{eqn | r = 0 \times z       | c = Ordinal Multiplication by Zero }} {{eqn | r = \left({x \times y}\right) \times z       | c = Ordinal Multiplication by Zero }} {{end-eqn}} === Case 2 === If $y \ne 0$, then $y \times z$ is a limit ordinal by Limit Ordinals Preserved Under Ordinal Multiplication. It follows that: {{begin-eqn}} {{eqn | l = x \times \left({y \times z}\right)       | r = \bigcup_{u \mathop < \left({y \times z}\right)} x \times u       | c = Definition of Ordinal Multiplication }} {{eqn | l = \left({x \times y}\right) \times z       | r = \bigcup_{w \mathop < z} \left({x \times y}\right) \times w       | c = Definition of Ordinal Multiplication }} {{end-eqn}} If $u < \left({y \times z}\right)$, then $u < \left({y \times w}\right)$ for some $w \in z$ by Ordinal is Less than Ordinal times Limit. {{begin-eqn}} {{eqn | l = x \times u       | o = \le       | r = x \times \left({y \times w}\right)       | c = Membership is Left Compatible with Ordinal Multiplication }} {{eqn | r = \left({x \times y}\right) \times w       | c = Inductive Hypothesis }} {{eqn | o = \le       | r = \left({x \times y}\right) \times z       | c = Subset is Right Compatible with Ordinal Multiplication }} {{end-eqn}} Generalizing, the result follows for all $u \in \left({y \times z}\right)$. Therefore by Supremum Inequality for Ordinals: :$x \times \left({y \times z}\right) \le \left({x \times y}\right) \times z$ Conversely, take any $w < z$. {{begin-eqn}} {{eqn | l = \left({x \times y}\right) \times w       | r = x \times \left({y \times w}\right)       | c = Inductive Hypothesis }} {{eqn | o = \le       | r = x \times \left({y \times z}\right)       | c = Membership is Left Compatible with Ordinal Multiplication }} {{end-eqn}} It follows by Supremum Inequality for Ordinals that: :$\left({x \times y}\right) \times z \le x \times \left({y \times z}\right)$ By definition of set equality: : $x \times \left({y \times z}\right) = \left({x \times y}\right) \times z$ This proves the limit case. {{qed}}