Ordinal arithmetics: $\alpha \cdot \omega \le \beta \implies \alpha + \beta = \beta$

Question

The original question was:

Let $\alpha,\beta$ be oridnals such that $\alpha\cdot\omega\leq\beta$.

prove/disprove: $\alpha+\beta=\beta$

So I tried to disprove with $\alpha=\omega,\beta=\omega^{\omega+1}$

But I got the feeling I did something wrong and I should actually prove that.

So I'll add a proof I thought of thanks to the comments:

$\alpha+\beta=\alpha+\alpha\cdot\omega+\gamma=\alpha(1+\omega)+\gamma=\alpha\cdot\omega+\gamma=\beta$

Is that correct?