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...

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.
