In this question, $\alpha$, $\beta$ and $\gamma$ are ordinals. I want to prove this by transfinite induction on $\gamma$, which typically has two or three cases. I'm considering three cases: the base case ($\gamma = 0$), the successor case ($\gamma = \delta + 1$) and the limit case.
I was struggling with the base case, which is so often just 'trivial'. If $\gamma = 0$, then this means that $\beta < \gamma = 0$, which doesn't make much sense to me.
An explanation of the base case would be greatly appreciated. Thanks.
In the base case $\gamma=0$, you want to prove the following statement:
This is vacuously true, since there are no ordinals $\beta<0$. For the statement to be false, there would need to exist a counterexample: there would need to exist $\alpha>1$ and $\beta<0$ such that $\alpha^\beta<\alpha^0$ is false. But no counterexample exists, since there does not exist any $\beta<0$ at all. Thus the statement is true.