Prove for an ordinal $\alpha\ne 0$, exists an ordinal $\epsilon$ such that $\omega^{\epsilon}\le \alpha\le \omega^{\epsilon+1}$
A hint is: Let $\theta$ be the smallest ordinal such that $\alpha<\omega^\theta$, prove that $\theta$ cannot be a limit ordinal.
I think once I prove that the least ordinal $\theta$ such that $\alpha<\omega^\theta$ is not a limit ordinal, then $\theta=\gamma^+$ some ordinal $\gamma$ as $\gamma\in\gamma^+$, by definition of "$<$" we have $\gamma<\theta$. Recall the fact that $\theta$ is the least ordinal such that $\alpha<\omega^{\theta}$, so $\alpha\ge \omega^\beta$ for any $\beta\le \theta$, in particular, $\alpha\ge \omega^{\gamma}$. So $\omega^\gamma\le \alpha\le \omega^{\gamma^+}$ as desired.
Suppose, in order to get a contradition, that $\theta$ is a non-zero limit ordinal($\theta$ cannot be zero since $\omega^0=1$ and $\alpha\ne 0$ so $\theta=0$ contradicts $\alpha<\omega^\theta$). I have stucked on proving that $\theta$ cannot be such an ordinal for a while... So could someone please help(on proving the hint)?
Thanks in advance.
A slightly expanded version of the hint(s) from the comments: keep in mind that if $\sigma$ is a limit ordinal, then we have $\sigma = \bigcup_{\tau\lt\sigma}\tau$; also, if $\sigma$ is a limit ordinal, then $\omega^\sigma$ is a limit ordinal with $\omega^\sigma = \bigcup_{\tau\lt\sigma}\omega^\tau$. Now, saying that $\alpha\lt\omega^\sigma$ is the same as saying that $\alpha\in\omega^\sigma$, and thus that $\alpha\in\bigcup_{\tau\lt\sigma}\omega^\tau$.
Now, what is the definition of $\alpha\in\bigcup_i S_i$? That is, under what conditions do we say that membership in a union holds?