So, I know that the operation "multiplication" in a group is not familiar old multiplication with real numbers. So, exponents in groups is not the same as exponents when dealing with real numbers.
I cannot assume $a^ma^n$=$a^{m+n}$, nor $(a^m)^n=a^{mn}$.
How do I go about setting up the proof for If a ∈ G and n ∈ N then both $(a^n)^{-1}$ and $(a^{-1})^n$ have unambiguous interpretations in terms of the definitions already given. Prove that these two are in fact equal.
Proof: (by Mathematical Induction)
[The Basis Step]
Let n = 1. Then, for n = 1, $(a^n)^{-1}$=$(a^1)^{-1}$=$a^{-1}$=$(a^{-1})^1$=$(a^{-1})^n$.
So, for n=1, $(a^n)^{-1}$=$(a^{-1})^n$.
[End of Basis Step]
[Inductive Step]
Let k ∈ N.
Suppose that $(a^k)^{-1}$=$(a^{-1})^k$.
Since $(a^k)^{-1}$=$(a^{-1})^k$,
then $(a^{(k+1)})^{-1}$
Well, for $n=1$, $(a^{-1})^1 = a^{-1}$ and $(a^1)^{-1}= a^{-1}$. So both sides are equal.
If $(a^{-1})^k= (a^k)^{-1}$ by induction hypothesis, then $(a^{k+1})^{-1} = (a^k a)^{-1} = a^{-1} (a^k)^{-1}$
where I have used that $(gh)^{-1}=h^{-1}g^{-1}$, and then
$a^{-1} (a^k)^{-1} = a^{-1} (a^{-1})^k$
by induction and then
$a^{-1} (a^{-1})^k = (a^{-1})^{k+1}.$
Moreover, I have used the definition of taking powers: $a^1=a$, $a^{k+1}=a^ka$ for $k\geq 1$.