= Solution
\b[False.] Take $\alpha=\omega_1+1$. Continuity of <ordinal multiplication> in the right argument gives $(\omega_1+1)\omega_1=\omega_1^2$. For a precise bound, for each countable ordinal $\beta$ monotonicity gives
$$
\omega_1\beta\leq(\omega_1+1)\beta\leq(\omega_1\cdot2)\beta
=\omega_1(2\beta).
$$
Both bounding expressions have supremum $\omega_1^2$ as $\beta$ ranges below $\omega_1$, because $2\beta$ is still countable and these indices are cofinal. But the successor rule gives
$$
\boxed{\alpha\omega_1=\omega_1^2,\qquad
\omega_1\alpha=\omega_1^2+\omega_1>\omega_1^2.}
$$
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