Can someone please tell me if my solution is okay?
Prove that for any $g\in G$ and $m,n\in \mathbb{Z}$, $g^{m}g^{n}=g^{m+n}$.
Case 1: $m,n>0$
$g^{m}g^{n}=g\cdot g\cdot \cdots \cdot g$ ($m$ $g$'s) $\cdot g \cdot g \cdot \cdots \cdot g$ ($n$ $g$'s) $=g\cdot g\cdot \cdots \cdot g$ ($m+n$ $g$'s) $=g^{m+n}$
Case 2: $m,n<0$
$g^{-m}g^{-n}=(g^{-1})^{m}(g^{-1})^{n}=g^{-1}\cdot g^{-1}\cdot \cdots \cdot g^{-1}$ ($m$ $g^{-1}$'s) $\cdot g^{-1} \cdot g^{-1} \cdot \cdots \cdot g^{-1}$ ($n$ $g^{-1}$'s) $=g^{-1}\cdot g^{-1}\cdot \cdots \cdot g^{-1}$ ($m+n$ $g^{-1}$'s) $=(g^{-1})^{m+n}=g^{-m-n}$
Case 3: $m>0$ and $n<0$
$g^{m}g^{-n}=g^{m}(g^{-1})^{n}=g\cdot g\cdot \cdots \cdot g$ ($m$ $g$'s) $\cdot g^{-1} \cdot g^{-1} \cdot \cdots \cdot g^{-1}$ ($n$ $g^{-1}$'s) $=g\cdot g\cdot \cdots \cdot g$ ($m-n$ $g$'s by cancellation laws) $=g^{m-n}$
Case 4: $m<0$ and $n>0$
$g^{-m}g^{n}=(g^{-1})^{m}g^{n}=g^{-1}\cdot g^{-1}\cdot \cdots \cdot g^{-1}$ ($m$ $g^{-1}$'s) $\cdot g \cdot g \cdot \cdots \cdot g$ ($n$ $g$'s) $=g\cdot g\cdot \cdots \cdot g$ ($n-m$ $g$'s by cancellation laws) $=g^{n-m}=g^{-m+n}$
Your proof is "okay," certainly, but you haven't covered all your bases.
In particular, you leave out every case in which at least one of $m,n$ is equal to $0.$
As pointed out in the comments above, there are those (perhaps including the person grading your assignment) who will (in my opinion, unnecessarily) insist on induction being used in the proof. I will agree that induction makes the proof easier, but I won't insist that it's necessary. However, I will admit that "$\dots$" gets in the way, here. It would be better to put it into words. For example, we might instead say (in case $1$): "$g^mg^n$ is simply the product of $m$ $g$'s and $n$ $g$'s, or of $m+n$ $g$'s, which is simply $g^{m+n}.$"