= Solution
Suppose first that $\delta>2$ is <multiplicatively closed ordinal>[multiplicatively closed]. It is also additively closed. Indeed, for $\beta,\gamma<\delta$, let $\mu=\max\{\beta,\gamma\}$. Monotonicity of <ordinal addition> gives
$$
\beta+\gamma\leq\mu+\mu=\mu\,2<\delta,
$$
where the last inequality uses $\mu<\delta$, $2<\delta$, and multiplicative closure. Part (b) therefore gives
$$
\delta=\omega^\lambda
$$
for some nonzero ordinal $\lambda$.
For any $\rho,\sigma<\lambda$, strict monotonicity gives
$$
\omega^\rho,\omega^\sigma<\omega^\lambda=\delta.
$$
Multiplicative closure and the exponent law now imply
$$
\omega^{\rho+\sigma}
=\omega^\rho\omega^\sigma
<\omega^\lambda,
$$
hence $\rho+\sigma<\lambda$. Thus $\lambda$ is additively closed, and part (b) gives $\lambda=\omega^\alpha$. Consequently
$$
\boxed{\delta=\omega^{\omega^\alpha}}.
$$
Conversely, let $\delta=\omega^\lambda$ with $\lambda=\omega^\alpha$. By part (b), $\lambda$ is additively closed. Take nonzero $\beta,\gamma<\delta$, with leading exponents $\rho,\sigma<\lambda$ in <Cantor normal form>. If $\gamma$ is finite, the product $\beta\gamma$ has leading exponent $\rho<\lambda$. If $\gamma$ is infinite, the <leading exponent of an ordinal product> is
$$
\rho+\sigma<\lambda
$$
by additive closure of $\lambda$. In either case
$$
\beta\gamma<\omega^\lambda=\delta.
$$
Products involving zero are immediate, so $\delta$ is multiplicatively closed. This is the <multiplicative closure criterion for a power of omega> and completes both directions.
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