Prove $(a^m)^n=a^{mn}$ for all $a\in G$ and $m,n\in\mathbb{Z}$ I have to prove  $(a^m)^n=a^{mn}$ for all $a\in G$ and $m,n\in\mathbb{Z}$ where $G$ is a group. Is it enough to just expand $(a^m)^n=(a^m***a^m)$- $n$ times. And then from here we can expand it a bit more to there there are $mn$ amount of $a's$? Or do I need to break it up into cases. I felt if I did I'll have atleast 3 cases and a few subcases. As of right now I have two cases one where $m=n=0$ and the other where $a,b\neq 0$.    
Here is what I have so far. 
New proof #2

Case 1: Let $m>0$ and $n>0$ We will proceed by induction. We fix m and induct on n. 
  Base case: Let n=1. We see that $a^m=a^m$. 
  Inductive case: Suppose that $(a^m)^k=a^{mk}$ We shall prove $(a^m)^{k+1}=a^{m(k+1)}$.
  It follows immediately from assumption that $(a^m)^{k+1}=a^{m(k+1)}$.
Case 2: $m=n=0$. It is immediately obvious that  $(a^m)^n=a^{mn}$
Case 3: $m<0$ and $n<0$.  Let $m=-t$ and $n=-r$ where $t,r>0$. Then $(a^m)^n=(a^{-t})^{-r})=(a^{-1})^t)^r)^{-1}=(a^{-1})^{rt})^{-1}=(a^{-nt})^{-1}=(a^{nt})^{-1}=a^{n*(-1)t}=a^{mn}$

 A: 
Case 1: Let $m>0$ and $n>0$ We will proceed by induction. We fix m and induct on n. 
  Base case: Let n=1. We see that $a^m=a^m$. 
  Inductive case: Suppose that $(a^m)^k=a^{mk}$ We shall prove $(a^m)^{k+1}=a^{m(k+1)}$.
  It follows immediately from assumption that $(a^m)^{k+1}=a^{m(k+1)}$.
Case 2: $m=n=0$. It is immediately obvious that  $(a^m)^n=a^{mn}$
Case 3: $m<0$ and $n<0$.  Let $m=-t$ and $n=-r$ where $t,r>0$. Then $(a^m)^n=(a^{-t})^{-r})=(a^{-1})^t)^r)^{-1}=(a^{-1})^{rt})^{-1}=(a^{-nt})^{-1}=(a^{nt})^{-1}=a^{n*(-1)t}=a^{mn}$

A: This is more a comment
than an answer,
but
you nowhere say how you define
$a^n$.
You also do not say
if the group is Abelian.
My guess is that the definition is something like this:
$a^0 = e$
(the identity element).
If $n$ is an non-negative integer
such that
$a^n$ is defined,
then
$a^{n+1}
=a(a^n)
$
(or maybe
$(a^n)a$).
If $n < 0$,
then
$a^n
=(a^{-1})^{-n}
$.
To me,
this looks like the problem
of defining multiplication
in terms of addition.
I recommend
Landau's
"Foundations of Analysis"
for a bracingly rigorous
discussion of these and
similar matters.
