 Solution

Use transfinite induction on nonzero $\alpha$. Part (b) gives $\alpha ={\omega }^{\delta }n+\zeta$, with $n$ a positive natural number and $\zeta <{\omega }^{\delta }\le \alpha$.

If $\zeta =0$, this is already a one-term Cantor normal form. Otherwise $0<\zeta <\alpha$, so the induction hypothesis supplies a finite Cantor normal form for $\zeta$. Its leading exponent $\eta$ satisfies ${\omega }^{\eta }\le \zeta <{\omega }^{\delta }$, hence $\eta <\delta$. All later exponents are smaller still. Prefixing ${\omega }^{\delta }n$ therefore givesThe number of terms is finite because the inductive representation of the strictly smaller remainder is finite. Equivalently, repeated extraction cannot produce an infinite strictly descending sequence of ordinals. Since $\alpha \ne 0$, the number of terms is positive.