The proof shall proceed by finite induction on $n$: For all $n \in \N_{\le 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \sum_{i \mathop = 1}^n \left({ x^{a_i} b_i}\right) \times x^y = x^{a_1 \mathop + y}$ Since $x$ is a limit ordinal, it follows that $x^y$ is a limit ordinal by Limit Ordinals Closed under Ordinal Exponentiation. === Basis for the Induction === $P \left({1}\right)$ is the statement: :$\displaystyle \sum_{i \mathop = 1}^1 \left({x^{a_i} b_i}\right) \times x^y = x^{a_1 \mathop + y}$ {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^1 \left({ x^{a_i} b_i }\right) \times x^y       | r = \left({ x^{a_i} \times b_i }\right) \times x^y       | c = Definition of Ordinal Sum }} {{eqn | r = x^{a_i} \times \left({ b_i \times x^y }\right)       | c = Ordinal Multiplication is Associative }} {{eqn | r = x^{a_i} \times x^y       | c = Finite Ordinal Times Ordinal }} {{eqn | r = x^{a_i \mathop + y}       | c = Ordinal Sum of Powers }} {{end-eqn}} This is our basis for the induction. {{qed|lemma}} === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$\displaystyle \sum_{i \mathop = 1}^k \left({ x^{a_i} b_i }\right) \times x^y = x^{a_1 + y}$ Then we need to show: :$\displaystyle \sum_{i \mathop = 1}^{k \mathop + 1} \left({ x^{a_i} b_i }\right) \times x^y = x^{a_1 \mathop + y}$ === Induction Step === This is our induction step: By Upper Bound of Ordinal Sum, it follows that: :$\displaystyle \sum_{i \mathop = 1}^n \left({ x^{a_{i \mathop + 1} } b_{i \mathop + 1} }\right) < x^{a_1}$ By Membership is Left Compatible with Ordinal Multiplication: {{begin-eqn}} {{eqn | l = x^{a_1} b_1       | o = \le       | r = x^{a_1} b_1 + \sum_{i \mathop = 1}^k \left({ x^{a_{i \mathop + 1} } b_{i \mathop + 1} }\right)       | c = Ordinal is Less than Sum }} {{eqn | o = \le       | r = x^{a_1} b_1 + x^{a_1}       | c = Membership is Left Compatible with Ordinal Addition }} {{eqn | ll = \implies       | l = x^{a_1} \times b_1 \times x^y       | o = \le       | r = \sum_{i \mathop = 1}^{k \mathop + 1} \left({ x^{a_i} b_i}\right) \times x^y       | c = Subset is Right Compatible with Ordinal Multiplication and General Associative Law for Ordinal Sum }} {{eqn | o = \le       | r = x^{a_1} \times \left({ b_1 + 1 }\right) \times x^y       | c = Subset is Right Compatible with Ordinal Multiplication }} {{end-eqn}} But: {{begin-eqn}} {{eqn | l = x^{a_1} \times b_1 \times x^y       | r = x^{a_1} \times x^y       | c = Finite Ordinal Times Ordinal }} {{eqn | r = x^{a_1 \mathop + y}       | c = Ordinal Sum of Powers }} {{eqn | l = x^{a_1} \times \left({ b_1 + 1 }\right) \times x^y       | r = x^{a_1} \times x^y       | c = Finite Ordinal Times Ordinal }} {{eqn | r = x^{a_1 \mathop + y}       | c = Ordinal Sum of Powers }} {{end-eqn}} Therefore: {{begin-eqn}} {{eqn | l = x^{a_1 \mathop + y}       | o = \le       | r = \sum_{i \mathop = 1}^n \left({ x^{a_i} b_i }\right) \times x^y       | c = Substitutivity of Class Equality }} {{eqn | o = \le       | r = x^{a_1 \mathop + y}       | c = Substitutivity of Class Equality }} {{eqn | ll= \implies       | l = \sum_{i \mathop = 1}^n \left({ x^{a_i} b_i }\right) \times x^y       | r = x^{a_1 \mathop + y}       | c = Definition of Set Equality }} {{end-eqn}} {{qed}}