Suggestions on how to prove the following equality. $a^{m+n}=a^m a^n$ Let $a$ be a nonzero number and $m$ and $n$ be integers. Prove the following equality: $a^{m+n}=a^{m}a^{n}$
I'm not really sure what direction to go in. I'm not sure if I need to show for $n$ positive and negative separately or is there an easier way. Can you use induction on integers?
My attempt:
1) Base case $m=0$. Prove that $a^{m+n}=a^m a^n$. Is true.
2) Assume the result holds for $m$. 
So I want to prove it holds for $m+1$. 
So I know that $a^{m+n}=a^ma^n$.
So does this imply that: $a^{m+n+1}= a^{m+1+n}= a^{m+1}a^n$? Am I going in the right direction? I'm not sure what to do next...
 A: As I mention in the comment, you need some definition of $a^n$ in order to get started. It turns out that $a^n: \mathbb{R}\setminus\{0\} \times \mathbb{Z}\to \mathbb{R}$ is uniquely determined by specifying the relationships


*

*$a^1 = a$;

*$a^n = a\cdot a^{n-1}$.


(Can you see why both of these are necessary?)
As a first fact, $a^1 = a\cdot a^0$, which shows that for $a\neq 0$, $a^0 = 1$.
Now we can prove your theorem, for $n\geq 0$, by induction on $n$.
Base case: $n=0$. Then $a^{m+n} = a^{m+0} = a^m = a^m \cdot 1 = a^m a^0 = a^m a^n.$
Inductive case: suppose $a^{m+n} = a^ma^n.$ Then 
$$a^{m+(n+1)} = a^{(m+n)+1} = a\cdot a^{m+n} = a\cdot a^ma^n = a^m(a\cdot a^n) = a^m(a^{n+1}).$$
As mentioned in the comments, you now need to show the relation also holds for $n<0$. I'll let you try that one on your own for now.
EDIT: Notice that I lumped together many "obvious" steps involving commutativity and associativity of addition and multiplication, more than one should for a proof at this level. It is worthwhile going over the proof in very careful steps and noting where and how all of the axioms of arithmetic are used; this then gives you better understanding of when and how the theorem breaks down for e.g. matrix exponentiation.
A: You wanna show that $$a^{m+n}=a^{m}a^n$$ for $m,n\in\Bbb Z$. I will do the case $m,n\geq 0$. I assume you define $a^n$ recursively as $$a^0=1\\a^n=a\cdot a^{n-1}$$
Now, fix $n$. We induct on $m$. For $m=0$ the result is true, since $a^0:=1\;,\;n+0=n$.Thus, assume the result true for $m$; that is, assume $a^{m+n}=a^ma^n$ holds. Now $$a^{m+n+1}:=a\cdot a^{m+n}=^{*}a\cdot a^ma^n=a^{m+1}a^n $$
where in $*$ we used the inductive hypothesis. Since $n$ was arbitrary, the result follows.
