# Ordinal Exponentiation via Cantor Normal Form/Limit Exponents

## Theorem

Let $x$ and $y$ be ordinals.

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

Let $\sequence {a_i}$ be a sequence of ordinals that is strictly decreasing on $1 \le i \le n$.

Let $\sequence {b_i}$ be a sequence of natural numbers.

Then:

$\ds \paren {\sum_{i \mathop = 1}^n x^{a_i} \times b_i}^y = x^{a_1 \mathop \times y}$

## Proof

$\ds \sum_{i \mathop = 1}^n \paren {x^{a_i} \times b_i} \le x^{a_1} \times \paren {b_1 + 1}$

Furthermore:

$\ds x^{a_1} \le \sum_{i \mathop = 1}^n \paren {x^{a_i} \times b_i}$

It follows that:

 $\ds \paren {x^{a_1} }^y$ $\le$ $\ds \paren {\sum_{i \mathop = 1}^n x^{a_i} \times b_i}^y$ Subset is Right Compatible with Ordinal Exponentiation $\ds$ $\le$ $\ds \paren {x^{a_1} \times \paren {b_1 + 1} }^y$ Subset is Right Compatible with Ordinal Exponentiation $\ds$ $=$ $\ds x^{a_1 \times y}$ Ordinal Exponentiation of Terms

It follows that:

 $\ds x^{a_1 \mathop + y}$ $\le$ $\ds \paren {\sum_{i \mathop = 1}^n x^{a_i} b_i}^y$ Ordinal Power of Power $\ds$ $\le$ $\ds x^{a_1 \mathop + y}$ proven above $\ds \leadsto \ \$ $\ds \paren {\sum_{i \mathop = 1}^n x^{a_i} b_i}^y$ $=$ $\ds x^{a_1 \mathop + y}$ Definition 2 of Set Equality

$\blacksquare$