Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian. 
Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian.

This question is from group theory in Abstract Algebra and no matter how many times my lecturer teaches it for some reason I can't seem to crack it. 
(Please note that $e$ in the question is the group's identity.)
Here's my attempt though...
First I understand Abelian means that if $g_1$ and $g_2$ are elements of a group $G$ then they are Abelian if $g_1g_2=g_2g_1$...
So, I begin by trying to play around with the elements of the group based on their definition...
$$(g_2g_1)^r=e$$ 
$$(g_2g_1g_2g_2^{-1})^r=e$$
$$(g_2g_1g_2g_2^{-1}g_2g_1g_2g_2^{-1}...g_2g_1g_2g_2^{-1})=e$$
I assume that the $g_2^{-1}$'s and the $g_2$'s cancel out so that we end up with something like,
$$g_2(g_1g_2)^rg_2^{-1}=e$$
$$g_2^{-1}g_2(g_1g_2)^r=g_2^{-1}g_2$$
Then ultimately...
$$g_1g_2=e$$
I figure this is the answer. But I'm not totally sure. I always feel like I do too much in the pursuit of an answer when there's a simpler way.
Reference: Fraleigh p. 49 Question 4.38 in A First Course in Abstract Algebra. 

For any $g, h \in G$, consider the element $g\cdot h\cdot h\cdot g.~$
Since $g^2 = g\cdot g= e$ for all $g \in G$, we find that
$$g\cdot h\cdot h\cdot g = g\cdot(h\cdot h)\cdot g = g\cdot e\cdot g = g\cdot g = e.$$
But, $g\cdot h$ has unique inverse element $g\cdot h$, while we have just proved that $(g\cdot h)\cdot (h\cdot g) = e$, and so it must be that $g\cdot h = h\cdot g$ for all $g, h \in G$, that is, $G$ is an abelian group.
