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} $. 

I assume that you would like to use $(a^{-1})^n = a^{-n}$ by saying that if $n > 0$ then
$$
a^m a^{-n} = a^m (a^{-1})^n = a^{m-n}
$$
by leveraging the case for $m,n > 0$. The problem is that you can't really do this, because in that case you had two powers with the same basis, but usually $a \neq a^{-1}$.

Instead, what you could go this way. First observe that, up to swapping indices and "inverting everything", we can assume $m \geq 0$ and $m \geq n$.
Now, you already proved the case with $m,n$ positive, so let's consider $n \leq 0$. If $n$ or $m$ are $0$ the claim is clearly true (because $a^0 = e$). Otherwise assume $m > 0$ and suppose the claim true for $n$. To prove it for $n - 1$ observe that
$$
a^m a^{n-1} = (a^{m-1} a) (a^{-1} a^n) = a^{m-1} e a^n = a^{m-1+n}.
$$
