The proof shall proceed by Transfinite Induction on $z$. === Basis for the Induction === {{begin-eqn}} {{eqn|l = \left({x^y}\right)^0      |r = 1      |c = definition of ordinal exponentiation }} {{eqn|r = x^0      |c = definition of ordinal exponentiation }} {{eqn|l =       |r = x^{y \mathop \times 0}      |c = definition of ordinal multiplication }} {{end-eqn}} This proves the basis for the induction. {{qed|lemma}} === Induction Step === Suppose that $\left({x^y}\right)^z = x^{y \mathop \times z}$. Then: {{begin-eqn}} {{eqn|l = \left({x^y}\right)^{z^+}      |r = \left({x^y}\right)^z \times x^y      |c = definition of ordinal exponentiation }} {{eqn|r = x^{y \mathop \times z} \times x^y      |c = by Inductive Hypothesis }} {{eqn|r = x^{\left({y \mathop \times z}\right) + y}      |c = by Ordinal Sum of Powers }} {{eqn|r = x^{y \mathop \times z^+}      |c = definition of ordinal multiplication }} {{end-eqn}} This proves the induction step. {{qed|lemma}} === Limit Case === Suppose $\left({x^y}\right)^w = x^{y \mathop \times w}$ for all $w \in z$ where $z$ is a limit ordinal. Then: {{begin-eqn}} {{eqn|lo = \forall w \in z:      |l = \left({x^y}\right)^w      |r = x^{y \mathop \times w}      |c = by Inductive Hypothesis }} {{eqn|ll = \implies      |l = \bigcup_{w \mathop \in z} \left({x^y}\right)^w      |r = \bigcup_{w \mathop \in z} x^{y \mathop \times w}      |c = by Indexed Union Equality }} {{eqn|ll = \implies      |l = \left({x^y}\right)^z      |r = \bigcup_{w \mathop \in z} x^{y \mathop \times w}      |c = definition of ordinal exponentiation }} {{end-eqn}} To prove the statement, it is necessary and sufficient to prove that: :$\displaystyle \bigcup_{w \mathop \in z} x^{y \mathop \times w} = x^{y \mathop \times z}$ The proof of this shall proceed by cases: ==== Case 1 ==== If $y = 0$, it follows that: {{begin-eqn}} {{eqn|l = \left({x^y}\right)^z      |r = 1^z      |c = definition of ordinal exponentiation }} {{eqn|r = 1      |c = by Exponent Base of One }} {{eqn|r = x^0      |c = definition of ordinal exponentiation }} {{eqn|r = x^{y \mathop \times z}      |c = by Ordinal Multiplication by Zero }} {{end-eqn}} ==== Case 2 ==== If $y \ne 0$: {{begin-eqn}} {{eqn|l = w      |o = \in      |r = z      |c =  }} {{eqn|ll = \implies      |l = \left({y \mathop \times w}\right)      |o = \le      |r = \left({y \mathop \times z}\right)      |c = by Membership is Left Compatible with Ordinal Multiplication and Ordinal Multiplication by Zero }} {{eqn|ll = \implies      |l = x^{y \mathop \times w}      |o = \le      |r = x^{y \mathop \times z}      |c = by Membership is Left Compatible with Ordinal Exponentiation and definition of ordinal exponentiation }} {{end-eqn}} Therefore, by Supremum Inequality for Ordinals, it follows that: :$\bigcup_{w \mathop \in z} x^{y \mathop \times w} \le x^{y \mathop \times z}$ Conversely: {{begin-eqn}} {{eqn|l = v      |o = \in      |r = y \times z      |c =  }} {{eqn|ll = \implies      |lo = \exists w \in z:      |l = v      |o = \in      |r = y \times w      |c = by Ordinal is Less than Ordinal times Limit }} {{eqn|ll = \implies      |l = x^v      |o = \le      |r = x^{y \mathop \times w}      |c = by Membership is Left Compatible with Ordinal Exponentiation }} {{end-eqn}} By Supremum Inequality for Ordinals, it follows that: :$\displaystyle \bigcup_{v \mathop \in y \mathop \times z} x^v \le \bigcup_{w \mathop \in z} x^{y \mathop \times w}$ Therefore, by the definition of ordinal exponentiation: {{begin-eqn}} {{eqn|l = x^{y \mathop \times z}      |r = \bigcup_{w \mathop \in z} x^{y \mathop \times w}      |c =  }} {{eqn|r = \left({x^y}\right)^z      |c =  }} {{end-eqn}} This proves the limit case. {{qed}}