Exercise 18, § 2

In the definition procedure defined in Exercise 17 (b), take $\alpha_0 = 1 + 1$ (denoted by 2 by abuse of language),

Denote $f(\xi,\eta)$ by $\xi^\eta$ and define $\alpha^0$ to be 1 for all ordinals $\alpha$. Also define $0^\beta$ to be 0 and $1^\beta$ to be 1 for all ordinals $\beta\geq 1$.

(a) Show that if $\alpha>1$ and $\beta<\beta'$, we have $\alpha^\beta<\alpha^{\beta'}$ and that, for each ordinal $\alpha>1$, $\alpha^\xi$ is a normal functional symbol with respect to $\xi$. Moreover, if $0<\alpha\leq\alpha'$, we have $\alpha^\beta\leq{\alpha'}^\beta$.

(b) Show that $\alpha^\xi\cdot\alpha^\eta=\alpha^{\xi+\eta}$ and $(\alpha^\xi)^\eta=\alpha^{\xi\eta}$.

(c) Show that, if $\alpha\geq 2$ and $\beta\geq 1$, $\alpha^\beta\geq\alpha\beta$.

(d) For each pair of ordinals $\beta\geq 1$ and $\alpha\geq 2$, there exist three ordinals $\xi,\gamma,\delta$ such that $\beta=\alpha^\xi\gamma+\delta$, where $0<\gamma<\alpha$ and $\delta<\alpha^\xi$, and these three ordinals are uniquely determined by these conditions.