= Solution
Use <transfinite induction> on nonzero $\alpha$. Part (b) gives $\alpha=\omega^\delta n+\zeta$, with $n$ a positive <natural number> and $\zeta<\omega^\delta\leq\alpha$.
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 $\omega^\eta\leq\zeta<\omega^\delta$, hence $\eta<\delta$. All later exponents are smaller still. Prefixing $\omega^\delta n$ therefore gives
$$
\boxed{\alpha=\omega^{\delta_0}n_0+\cdots+\omega^{\delta_{k-1}}n_{k-1},\qquad
\delta_0>\cdots>\delta_{k-1},\quad0<n_j<\omega.}
$$
The 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 $\alpha\ne0$, the number of terms is positive.
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