Prove $(a^{-1}ba)^n = a^{-1}b^na$ for all $n \in \mathbb Z$ and $a, b$ in a group.
Assume $n \ge 1$. The identity is true for $n = 0, 1.$ Proof for $n + 1: (a^{-1}ba)^n = (a^{-1}ba)^{n + 1} = (a^{-1}ba)^n(a^{-1}ba) = (a^{-1}b^na)(a^{-1}ba) = a^{-1}b^na$
Assume $n \le -1.$ Since $-n \ge 1,$ we have $(a^{-1}ba)^n = [(a^{-1}ba)^{-1}]^{-n} = (a^{-1}b^{-1}a)^{-n} = a^{-1}({b^{-1}})^{-n}a = a^{-1}b^na.$
How are we inducting on $n$ in the second part of the proof for negative integers? Are we doing $n - 1$ or $-n - 1$?
Also is $[(a^{-1}ba)^{-1}]^{-n} = (a^{-1}b^{-1}a)^{-n} $ by inductive hypothesis?