 Solution

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 givesBoth 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