Lemma about ordinal arithmetic I'm trying to prove the following lemma:
For any three ordinals $\alpha, \alpha' >0$, $\beta >1$, if $\alpha > \alpha'$, then we have
$$
\beta^\alpha > \beta^{\alpha'} \cdot \delta + \gamma
$$
For $0 < \delta < \beta$ and $\gamma < \beta^{\alpha'}$. 
My question is, is this even true, and if so what is the best way to prove it?

Just note that $$\beta^\alpha\geq\beta^{\alpha'+1}=\beta^{\alpha'}\cdot \beta\geq\beta^{\alpha'}\cdot (\delta+1)=\beta^{\alpha'}\cdot\delta+\beta^{\alpha'}>\beta^{\alpha'}\cdot\delta+\gamma.$$
