The proof shall proceed by Transfinite Induction on $z$. === Basis for the Induction === :$x^0 = 1$ for all $x$ {{begin-eqn}} {{eqn|l = x^y \times x^z      |r = x^y      |c = by Ordinal Multiplication by Zero }} {{eqn|l =       |r = x^{y + z}      |c = by Ordinal Addition by Zero }} {{end-eqn}} This proves the basis for the induction. {{qed|lemma}} === Induction Step === Suppose that $x^y \times x^z = x^{y + z}$. Then: {{begin-eqn}} {{eqn|l = x^y \times x^{z^+}      |r = x^y \times \left({ x^z \times x }\right)      |c = definition of ordinal exponentiation }} {{eqn|l =       |r = \left({ x^y \times x^z }\right) \times x      |c = by Ordinal Multiplication is Associative }} {{eqn|l =       |r = x^{y + z} \times x      |c = by Inductive Hypothesis }} {{eqn|l =       |r = x^{\left({y + z}\right)^+}      |c = definition of ordinal exponentiation }} {{eqn|l =       |r = x^{\left({y + z^+}\right)}      |c = definition of ordinal addition }} {{end-eqn}} This proves the induction step. {{qed|lemma}} === Limit Case === Suppose that $\forall w \in z: x^y \times x^w = x^{y + w}$ for limit ordinal $z$. {{begin-eqn}} {{eqn|ll = \forall w \in z:      |l = \left({x^y \times x^w}\right)      |o = \le      |r = \left({x^y \times x^z}\right)      |c = by Membership is Left Compatible with Ordinal Multiplication }} {{eqn|ll = \implies      |l = \bigcup_{w \in z} \left({ x^y \times x^w }\right)      |o = \le      |r = \left({ x^y \times x^z }\right)      |c = by Indexed Union Subset }} {{end-eqn}} Conversely: {{begin-eqn}} {{eqn|l = w \in \left({ x^y \times x^z }\right)      |o = \implies      |r = \exists u \in x^z: w \in \left({ x^y \times u }\right)      |c = by Ordinal is Less than Ordinal times Limit }} {{eqn|o = \implies      |r = \exists v \in z: \exists u \in x^v: w \in \left({ x^y \times u }\right)      |c = by Ordinal is Less than Ordinal to Limit Power }} {{end-eqn}} But this means that $u$ is bounded above by $x^v$ for some $v \in z$. Thus there exists a $v \in z$ such that: :$w \le \left({ x^y \times x^v }\right)$ By Supremum Inequality for Ordinals, it follows that: :$\left({ x^y \times x^z }\right) \le \bigcup_{w \in z} \left({ x^y \times x^w }\right)$ {{begin-eqn}} {{eqn|l = \left({ x^y \times x^z }\right)      |r = \bigcup_{w \in z} \left({ x^y \times x^w }\right)      |c = Definition of Set Equality }} {{eqn|l =       |r = \bigcup_{w \in z} x^{y+w}       |c = Inductive Hypothesis for the Limit Case }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn|ll = \forall w \in z:      |l = y + w      |o = \le      |r = y + z      |c = by Membership is Left Compatible with Ordinal Addition }} {{eqn|ll = \forall w \in z:      |l = x^{y + w}      |o = \le      |r = x^{y + z}      |c = by Membership is Left Compatible with Ordinal Exponentiation }} {{eqn|ll = \implies      |l = \bigcup_{w \in z} x^{y + w}      |o = \le      |r = x^{y + z}      |c = by Indexed Union Subset }} {{end-eqn}} Conversely: {{begin-eqn}} {{eqn|ll =       |l = w \in x^{y + z}      |o = \implies      |r = \exists u \in \left({y + z}\right): w \in x^u      |c = definition of ordinal exponentiation }} {{eqn|ll =       |l =       |o = \implies      |r = \exists v \in z: \exists u \in \left({y + v}\right): w \in x^u      |c = definition of ordinal addition }} {{end-eqn}} Thus, $u$ is bounded above by $\left({ y + v }\right)$ for some $v \in z$. Therefore: :$x^u \le x^{y + v}$ By Supremum Inequality for Ordinals, it follows that: :$x^{y + z} \le \bigcup_{w \in z} x^{y + w}$ Thus, by definition of set equality: :$x^{y + z} = \bigcup_{w \in z} x^{y + w}$ {{qed|lemma}} Combining the results of the first and second lemmas for the limit case: :$x^{y+z} = x^y \times x^z$ This proves the limit case. {{qed}}