= Solution
<Ordinal exponentiation> is defined by
$$
\omega^0=1,
\qquad
\omega^{\beta+1}=\omega^\beta\omega,
\qquad
\omega^\lambda=\sup_{\beta<\lambda}\omega^\beta
$$
for limit $\lambda$. <Transfinite induction> gives $\omega^\alpha\geq\alpha$. Since $\omega^{\alpha+1}>\alpha$, there is a least $\alpha^*$ with $\omega^{\alpha^*}>\alpha$. If nonzero $\alpha^*$ were a limit, the defining supremum would imply $\omega^\beta>\alpha$ for some $\beta<\alpha^*$, a contradiction. Hence $\alpha^*$ is a successor.
For $\alpha>0$, write $\alpha^*=\beta+1$. Then $\omega^\beta\leq\alpha<\omega^{\beta+1}$. There is a largest positive integer $n$ with $\omega^\beta n\leq\alpha$, and ordinal division gives
$$
\alpha=\omega^\beta n+\gamma,
\qquad
\gamma<\omega^\beta.
$$
Repeating on $\gamma$ terminates because there is no infinite strictly decreasing sequence of ordinals, producing the <Cantor normal form>
$$
\alpha=\omega^{\beta_1}n_1+\cdots+\omega^{\beta_k}n_k,
\qquad
\beta_1>\cdots>\beta_k.
$$
For two monomials,
$$
\omega^{\delta_1}m_1+\omega^{\delta_2}m_2=
\begin{cases}
\omega^{\delta_2}m_2,&\delta_1<\delta_2,\\
\omega^{\delta_1}(m_1+m_2),&\delta_1=\delta_2,\\
\omega^{\delta_1}m_1+\omega^{\delta_2}m_2,&\delta_1>\delta_2.
\end{cases}
$$
For general $\alpha+\alpha'$, compare the leading exponent $\beta'_1$ with the exponents of $\alpha$: discard every trailing term of $\alpha$ having exponent below $\beta'_1$, combine coefficients if the last retained exponent equals $\beta'_1$, and then append all remaining terms of $\alpha'$. This is its Cantor normal form.
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