Let $\alpha, \beta, \gamma $ ordinal numbers . Prove $\alpha \leq \beta \Rightarrow \alpha^{\gamma} \leq \beta^{\gamma}$.
Attempt Induction under $\gamma$.
If $\gamma=0 \Rightarrow \alpha^{0} = 1 = \beta^{0}$ [but what happens if $\alpha = 0$ or $\beta =0$]
If $\exists \delta$ ordinal number such that $\delta +1= \gamma \Rightarrow \alpha^{\gamma}=\alpha^{\delta+1}= \alpha^{\delta}* \alpha \leq \beta^{\delta}*\alpha \leq \beta^{\delta}*\beta= \beta^{\delta+1}= \beta^{\gamma}$
If $\gamma \neq 0$ is a limit ordinal $\Rightarrow \forall \delta < \gamma, \alpha^{\delta} \leq \beta^{\delta} \Rightarrow \alpha^{\gamma}= \mbox{sup }\{\alpha^{\delta}: \delta < \gamma\} \leq \mbox{sup }\{\beta^{\delta}: \delta < \gamma\}= \beta^{\gamma}$
By principle of transfinite induction, $\forall \alpha, \beta, \gamma \mbox{ with } \alpha \leq \beta \Rightarrow \alpha^{\gamma} \leq \beta^{\gamma}$
Could you check, if my prove is correct?