# How to calculate a power of a sum of ordinals (ordinal binomial theorem?)

Hello,

Suppose we have a black box which allows us to compute $\alpha^\beta$ provided $\alpha$ has the form $\omega^{\gamma}$. And suppose we have a black box for decomposing an arbitrary ordinal into its Cantor Normal Form.

How can we use this to compute arbitrary ordinal exponents?

The problem obviously reduces to the case when the exponent itself is of the form $\omega^{\gamma}$ ($\gamma>0$). So the question is, how to compute $(\omega^{\gamma_1}+\omega^{\gamma_2}+\cdots+\omega^{\gamma_n})^{\omega^{\gamma}}$, provided $\gamma_1\geq\cdots\geq\gamma_n$ and $\gamma>0$?

$\omega^{\alpha\cdot \omega^\gamma} = (\omega^\alpha )^{\omega^\gamma} \le (\omega^\alpha + \cdots )^{\omega^\gamma} \le (\omega^{\alpha+1})^{\omega^\gamma} = \omega^{(\alpha+1)\cdot \omega^\gamma}$.
Wlog $\alpha,\gamma > 0$. Note that $2\cdot \omega^\gamma = \omega^\gamma$ for all $\gamma>0$, by induction and the associative law.
Now $(\alpha+1) \omega^\gamma \le \alpha\cdot 2 \cdot \omega^\gamma = \alpha\cdot \omega^\gamma$, and the result is $\omega^{\alpha \cdot \omega^\gamma}$.