 Solution

True. For $\beta >0$, multiplication $\gamma \mapsto \beta \gamma$ is strictly increasing and continuous at limit ordinals, and $\beta \gamma \ge \gamma$. Thus there is a greatest $\gamma$ with $\beta \gamma \le \alpha$: the set is bounded, and its supremum still satisfies the inequality by continuity if it is a limit. The remaining ordered interval in $\alpha$ has an ordinal order type $\delta$, so $\alpha =\beta \gamma +\delta$. If $\delta \ge \beta$, then $\beta (\gamma +1)\le \alpha$, contradicting maximality. Hence $\delta <\beta$.