= Solution
For a nonzero <ordinal> $\alpha$, let
$$
S=\{\xi:\omega^\xi\leq\alpha\}.
$$
The set is nonempty because $0\in S$, and it is an initial segment because <ordinal exponentiation> by $\omega$ is strictly increasing. Put $\beta=\sup S$.
If $\beta$ is a successor, the definition of the <supremum> forces $\beta\in S$. If $\beta$ is a <limit ordinal>, then continuity of ordinal exponentiation gives
$$
\omega^\beta
=\sup_{\xi<\beta}\omega^\xi
\leq\alpha,
$$
so again $\beta\in S$. Thus $\omega^\beta\leq\alpha$, while $\omega^{\beta+1}>\alpha$ by the definition of the supremum. Hence $\beta$ is the greatest required exponent.
Apply <division by an additively indecomposable ordinal> with $\lambda=\omega^\beta$:
$$
\alpha=\omega^\beta q+\gamma,
\qquad
\gamma<\omega^\beta.
$$
Since $\alpha\geq\omega^\beta$, the quotient $q$ is nonzero. It must be finite: if $q\geq\omega$, then
$$
\omega^{\beta+1}
=\omega^\beta\omega
\leq\omega^\beta q
\leq\alpha,
$$
contradicting the maximality of $\beta$. Writing $q=n$ gives
$$
\boxed{\alpha=\omega^\beta n+\gamma},
\qquad
0<n<\omega,\quad \gamma<\omega^\beta.
$$
This is the <greatest power of omega below an ordinal>[leading-term decomposition].
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