Ordinal Multiplication via Cantor Normal Form/Infinite Exponent

Theorem
Let $x$ and $y$ be ordinals.

Let $x > 1$ and let $y \le \omega$ where $\omega$ denotes the minimal infinite successor set.

Let $\langle a_i \rangle$ be a sequence of ordinals that is strictly monotone decreasing on $1 \le i \le n$.

Let $\langle b_i \rangle$ be a sequence of ordinals such that $0 < b_i < x$ for all $1 \le i \le n$.

Then:


 * $\displaystyle \sum_{i \mathop = 1}^n \left({ x^{a_i} b_i }\right) \times x^y = x^{a_1 \mathop + y}$

Proof
It follows that:

By multiplying the inequalities by $x^y$ on the left:

Solving for both sides of the inequality:

Therefore:

Also see

 * Cantor Normal Form