Question

For any elements a and to from a group and any integer n, prove that (a^-1ba)^n=a^-1b^na.

Exponential growth and decay
For any elements a and to from a group and any integer n, prove that $$\displaystyle{\left({a}^{{-{{1}}}}{b}{a}\right)}^{{n}}={a}^{{-{{1}}}}{b}^{{n}}{a}$$.

This is a direct computation: $$\displaystyle{\left({a}^{{-{{1}}}}{b}{a}\right)}^{{n}}={\left({a}^{{-{{1}}}}{b}{a}\right)}{\left({a}^{{-{{1}}}}{b}{a}\right)}\ldots{\left({a}^{{-{{1}}}}{b}{a}\right)}\rbrace{n}$$ times
$$=a^{-1}b(aa^{-1})b(aa^-1)...ba =a^-1bebe...ba =a^-1(bb*b)n \times)a =a^{-1}b^na$$