if $\gamma=\omega^\alpha$ for some ordinal $\alpha$, then $\gamma$ is indecomposable (explanation)

Question

I am reading the following proof, it shows that if $\gamma=\omega^\alpha$ for some ordinal $\alpha$, then $\gamma$ is indecomposable (e.i. $\beta_1+\beta_2<\gamma$ with $\beta_1, \beta_2<\gamma$):

Let $\beta_1, \beta_2<\gamma=\omega^\alpha$. there exist $\alpha^{\prime}<\alpha$ and $n<\omega$ such that $\beta_1, \beta_2<\omega^{\alpha^{\prime}} \cdot n$; consider Cantor's normal forms of $\beta_1, \beta_2$. Hence $\beta_1+\beta_2<\omega^{\alpha^{\prime}} \cdot(k+k)<\omega^\alpha=\gamma$

I don't understand why:

1. there exist $\alpha^{\prime}<\alpha$

2. $\beta_1, \beta_2<\omega^{\alpha^{\prime}} \cdot n$

3. $\beta_1+\beta_2<\omega^{\alpha^{\prime}} \cdot(k+k)$

could someone please clarify these questions for me?

Answer 1

For questions 1 & 2: Recall that $$\omega^\alpha = \begin{cases} \lim_{n<\omega} \omega^{\alpha'}\cdot n &\text{ if } \alpha = {\alpha'}+ 1 \\ \lim_{{\alpha'}<\alpha} \omega^{\alpha'}&\text{ if } \alpha \text{ is a limit} \end{cases}$$

Now, $\beta < \omega^\alpha$ implies $\beta$ is smaller than one of the elements in the limit; in both cases, it has the form $\omega^{\alpha'}\cdot n$.

For questions 3: I suspect it should have been $\beta_1+\beta_2< \omega^{\alpha^{\prime}} \cdot n + \omega^{\alpha^{\prime}} \cdot n=\omega^{\alpha^{\prime}} \cdot(n+n)<\omega^\alpha=\gamma$.