Write down the inductive definition of ordinal exponentiation. Show that $\omega^{\alpha} \geqslant \alpha$ for every ordinal $\alpha$. Deduce that, for every ordinal $\alpha$, there is a least ordinal $\alpha^{*}$ with $\omega^{\alpha^{*}}>\alpha$. Show that, if $\alpha \neq 0$, then $\alpha^{*}$ must be a successor ordinal.

Now let $\alpha$ be a non-zero ordinal. Show that there exist ordinals $\beta$ and $\gamma$, where $\gamma<\alpha$, and a positive integer $n$ such that $\alpha=\omega^{\beta} n+\gamma$. Hence, or otherwise, show that $\alpha$ can be written in the form

$$\alpha=\omega^{\beta_{1}} n_{1}+\omega^{\beta_{2}} n_{2}+\cdots+\omega^{\beta_{k}} n_{k}$$

where $k, n_{1}, n_{2}, \ldots, n_{k}$ are positive integers and $\beta_{1}>\beta_{2}>\cdots>\beta_{k}$ are ordinals. [We call this the Cantor normal form of $\alpha$, and you may henceforth assume that it is unique.]

Given ordinals $\delta_{1}, \delta_{2}$ and positive integers $m_{1}, m_{2}$ find the Cantor normal form of $\omega^{\delta_{1}} m_{1}+\omega^{\delta_{2}} m_{2}$. Hence, or otherwise, given non-zero ordinals $\alpha$ and $\alpha^{\prime}$, find the Cantor normal form of $\alpha+\alpha^{\prime}$ in terms of the Cantor normal forms

$$\alpha=\omega^{\beta_{1}} n_{1}+\omega^{\beta_{2}} n_{2}+\cdots+\omega^{\beta_{k}} n_{k}$$

and

$$\alpha^{\prime}=\omega^{\beta_{1}^{\prime}} n_{1}^{\prime}+\omega^{\beta_{2}^{\prime}} n_{2}^{\prime}+\cdots+\omega^{\beta_{k^{\prime}}^{\prime}} n_{k^{\prime}}^{\prime}$$

of $\alpha$ and $\alpha^{\prime}$.