I'm trying to show whenever $β<ω^γ$, $β$ has to be zero. The plan is that if $γ_0=γ$ is a limit ordinal, then $ω^{γ_0}$ being sup$\{ω^ε|ε<γ_0\}$, $β<ω^{γ_0}$ entails there exists $γ_1<γ_0$ such that $β<ω^{γ_1}$. If $γ_1$ is again a limit, then we can similarly find $γ_2<γ_1$ such that $β<ω^{γ_2}$, etc. If we never run into any non-limit $γ_i$, this process constructs a strictly decreasing sequence of ordinals, $γ_0>γ_1>γ_2>\cdots$, which must terminate at zero. Since $β<ω^{γ_i}$ for all $i$, we have $β<ω^0=1$ as expected.
To complete the proof, however, still we should address what if we do encounter $β<ω^{γ_i}$ where $γ_i$ is not a limit, say, $γ_i=ε+1$. If $β<ω^ε$, it's fine. If $β\geqω^ε$, I was able to convince myself that $β+ω^{γ_i}=ω^{γ_i}$. This is because (i) $β+ω^{γ_i}\geqω^ε+ω^{γ_i}=ω^{γ_i}$; (ii) dividing $β$ by $ω^ε$ as $β=ω^εx+y$, where $x<ω$, $y<ω^ε$, it follows $β<ω^εx+ω^ε$, so $β+ω^{γ_i}\leqω^εx+ω^ε+ω^{γ_i}=ω^εx+ω^{γ_i}=ω^ε(x+ω)=ω^εω=ω^{γ_i}$ (addition is non-decreasing in the first operand).
Now I'm stuck at this point. I'll appreciate if someone can explain how to proceed on, or maybe the proof plan is wrong from the beginning...
Update
Yes, I realized the plan wasn't on the right track. The statement boils down to two claims, (i) neither of $\alpha$ and $\beta$ exceed $ω^γ$, but (ii) they can't be both smaller than $ω^γ$ (as Deedlit mentions). (ii) follows from a simple fact that $β+ω^γ=ω^γ$ whenever $β<ω^γ$. Thanks for the help!
The case $\gamma = 0$ is trivial, so assume $\gamma > 0$.
If $\alpha + \beta = \omega^{\gamma}$ and $\beta \not \in \{0, \omega^{\gamma} \}$ then $\alpha < \omega^{\gamma} \wedge \beta < \omega^{\gamma}$. Hence $$ \begin{align*} \alpha + \beta &\overset{(\dagger)}{<} \alpha + \omega^\gamma \\ &\overset{(\ddagger)}= \sup \{ \alpha + \delta \mid \delta < \omega^\gamma \} \\ &\overset{\alpha < \omega^\gamma}{=} \sup \{ \delta \mid \delta < \omega^\gamma \} \\ &= \omega^\gamma \end{align*} $$ Contradiction!
$(\dagger)$ holds because ordinal addition is striclty increasing in the second coordinate.
$(\ddagger)$ holds because, for $\gamma \neq 0$, $\omega^\gamma$ is a limit ordinal.