In this paper, page $12,$ the authors quoted the following:
For a general nonzero countable ordinal $\beta_0,$ write $\beta_0$ in Cantor normal form as $$\beta_0 = \omega^{\beta_1}\cdot m_1 + ... + \omega^{\beta_k}\cdot m_k$$ where $k,m_1,...,m_k\in\mathbb{N}, \omega_1>\beta_1>\beta_2>...>\beta_k.$ If $\gamma_0\neq 0,$ then $\beta_0 \cdot \omega^{\gamma_0} = \omega^{\beta_1} \cdot \omega^{\gamma_0}.$
Question: How to obtain $\beta_0 \cdot \omega^{\gamma_0} = \omega^{\beta_1} \cdot \omega^{\gamma_0}$?
I thought it should be $$\beta_0\cdot\omega^{\gamma_0} = \omega^{\beta_1} \cdot \omega^{\gamma_0} + ...+\omega^{\beta_k} \cdot \omega^{\gamma_0}.$$
But they look different.
As has been discussed in the comments, the key observation is that for any $\alpha<\omega^{\beta_1}$, $\alpha+\omega^{\beta_1}=\omega^{\beta_1}$. It follows that \begin{align*} \beta_0\cdot 2 &= \omega^{\beta_1}\cdot m_1+(\omega^{\beta_2}\cdot m_2+\dots+\omega^{\beta_k}\cdot m_k+\omega^{\beta_1})+\omega^{\beta_1}\cdot(m_1-1)+\omega^{\beta_2}\cdot m_2+\dots+\omega^{\beta_k}\cdot m_k \\ &= \omega^{\beta_1}\cdot m_1+\omega^{\beta_1}+\omega^{\beta_1}\cdot(m_1-1)+\omega^{\beta_2}\cdot m_2+\dots+\omega^{\beta_k}\cdot m_k \\ &= \omega^{\beta_1}\cdot m_1\cdot 2+\omega^{\beta_2}\cdot m_2+\dots+\omega^{\beta_k}\cdot m_k. \end{align*}
In the same way, we can prove by induction on $\alpha$ that $$\beta_0\cdot\alpha=\omega^{\beta_1}\cdot m_1\cdot\alpha+\omega^{\beta_2}\cdot m_2+\dots+\omega^{\beta_k}\cdot m_k$$ if $\alpha$ is a successor and $$\beta_0\cdot\alpha=\omega^{\beta_1}\cdot m_1\cdot\alpha$$ if $\alpha$ is a limit. In particular, taking $\alpha=\omega^{\gamma_0}$, we find $$\beta_0\cdot\omega^{\gamma_0}=\omega^{\beta_1}\cdot m_1\cdot \omega^{\gamma_0}=\omega^{\beta_1}\cdot\omega^{\gamma_0}$$ (the latter equality is because $m_1\cdot \omega^{\gamma_0}=\omega^{\gamma_0}$).