 Solution

The ordinal operations are defined by transfinite recursion in the right argument. Addition satisfies $\alpha +0=\alpha$, $\alpha +(\beta +1)=(\alpha +\beta )+1$, and $\alpha +\lambda ={\sup }_{\beta <\lambda }(\alpha +\beta )$ at a nonzero limit. Multiplication satisfies $\alpha \cdot 0=0$, $\alpha \cdot (\beta +1)=\alpha \cdot \beta +\alpha$, and $\alpha \cdot \lambda ={\sup }_{\beta <\lambda }\alpha \cdot \beta$. Exponentiation satisfies ${\alpha }^0=1$, ${\alpha }^{\beta +1}={\alpha }^{\beta }\cdot \alpha$, and ${\alpha }^{\lambda }={\sup }_{\beta <\lambda }{\alpha }^{\beta }$ at nonzero limits. These are ordinal operations, not commutative cardinal arithmetic.

For a normal function on an ordinal, prove domination by transfinite induction. At zero, $0\le \Phi (0)$ automatically. If $\alpha \le \Phi (\alpha )$, strict increase at successors gives $\alpha +1\le \Phi (\alpha )+1\le \Phi (\alpha +1)$. At a nonzero limit $\lambda$, continuity and the induction hypotheses giveTherefore $\boxed{\alpha \le \Phi (\alpha )\text{ for every ordinal }\alpha }$. No axiom of choice is needed for this ordinal induction.