Counterexamples for $\alpha < \beta \Longrightarrow \alpha^{\gamma} < \beta^{\gamma}$, where $\alpha$, $\beta$ and $\gamma$ are nonzero ordinals I am trying to find counterexamples for the following formula:

$\alpha < \beta \Longrightarrow \alpha^{\gamma} < \beta^{\gamma}$, where $\alpha$, $\beta$ and $\gamma$ are nonzero ordinals.

I think $\gamma$ may have to be a limit ordinal, for instance, maybe $\omega$ (another notation for $\mathbb{N}$). Can anyone provide hints?
 A: I came up with this proof thanks to hints provided. Consider the following two ordinals: $\boldsymbol{2}^{\omega}$ and $\boldsymbol{3}^{\omega}$. The two ordinals can be expanded as
\begin{equation*}
\boldsymbol{2}^{\omega} = \bigcup\left\{\boldsymbol{2}^{\theta}: \theta \in \omega\right\},
\end{equation*}
and
\begin{equation*}
\boldsymbol{3}^{\omega} = \bigcup\left\{\boldsymbol{3}^{\theta}: \theta \in \omega\right\}.
\end{equation*}
It is obvious that $\omega$ is an upper bound for both $\left\{\boldsymbol{2}^{\theta}: \theta \in \omega\right\}$ and $\left\{\boldsymbol{3}^{\theta}: \theta \in \omega\right\}$.
Next assume that there exists some ordinal $\lambda$ such that $\lambda < \omega$ and that $\lambda$ is also an upper bound for $\left\{\boldsymbol{2}^{\theta}: \theta \in \omega\right\}$. Immediately we have $\lambda < \boldsymbol{2}^{\lambda} \in \left\{\boldsymbol{2}^{\theta}: \theta \in \omega\right\}$, contradicting that $\lambda$ is an upper bound of $\left\{\boldsymbol{2}^{\theta}: \theta \in \omega\right\}$. Thus, $\omega = \operatorname{lub}\left(\left\{\boldsymbol{2}^{\theta}: \theta \in \omega\right\}\right)$. Similarly, $\omega = \operatorname{lub}\left(\left\{\boldsymbol{3}^{\theta}: \theta \in \omega\right\}\right)$. Then we have
\begin{equation*}
\begin{aligned}
\boldsymbol{2}^{\omega} &= \bigcup\left\{\boldsymbol{2}^{\theta}: \theta \in \omega\right\}\\
&= \operatorname{lub}\left(\left\{\boldsymbol{2}^{\theta}: \theta \in \omega\right\}\right)\\
&= \omega\\
&= \operatorname{lub}\left(\left\{\boldsymbol{3}^{\theta}: \theta \in \omega\right\}\right)\\
&= \bigcup\left\{\boldsymbol{3}^{\theta}: \theta \in \omega\right\}\\
&= \boldsymbol{3}^{\omega}.
\end{aligned}
\end{equation*}
