Why does this inverse work $U_n$ is defined to be a set such that $ \{\ [a] \in \mathbb{Z_n} \mid \gcd(a,n)=1 \}\ $ where $ \{\ [0] \}\ $ is obviously omitted.
I'm proving that $(U_n , \cdot_n)$ is a group and I have come to the inverse part.
So I have noticed that $a^2 = 1, \ \ \ \forall \ [a] \in G$ 
The tricky part is proving this. Now I am not required to bring the most convoluted and rigorous proof in the world but a solid simple proof would be great.
I also realize that I may be wrong and that $a^2$ for any $a$ is not an inverse function in $G$. If that is the case, please throw me on the right track.

It's not true that $[a]^2 = [1]$ for all $[a]$ in all $\mathbb Z_n$.  For example $[2]^2 = [4]$ in $\mathbb Z_5$.
Hint: Let $[a] \in \mathbb Z_n$ with $(a, n) = 1$.  Did you know that if $(a, n) = 1$ then there exist integers $x, y$ such that $ax + ny = 1$?  What happens when you reduce that equation modulo $n$?
