Show that $a^{m} a^{n} = a^{m+n}$ Prove that if $G$ is a group and $a\in G$, then we have $\forall m,n\in\mathbb{Z} $ that 
$$a^{m} a^{n} = a^{m+n}.$$
I've proved the case when $m,n>0$ but I'm stuck on how to prove the case when one or both are negative without assuming that $(a^{-1})^n = a^{-n} $. 

"without assuming that $(a^{-1})^n=a^{−n}$". 
How was $a^{-n}$ (n positive) originally defined (if it was)?  As notation it doesn't have any intrinsic meaning.  It's very easy to show $(a^{-1})^n = (a^n)^{-1}$ (multiply either by $a^n$ and the result is $e$ so both of them are the unique inverse of $a^n$).  So it's reasonable and well defined to define $a^{-n}$ as either one.
But while you're at it, you also have to define $a^0 = e$.  
