How do you prove the group (Zn\{0}, *) has an inverse element?

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I understand that I want first suppose there is an $a$ and find a general $a^{-1}$. So I suppose that $[a]_n * [a^{-1}]_n = e = 1$.

I'm struggling to find $a^{-1}$.

I think I need to I want to say that $a^{-1}$ is $1/[a]_n$, but then the inverse will not be an integer, which is absurd. Or maybe I try to find the order of a where it will become $e$?

Thanks.

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The usual proof that an element $a \in \mathbb{Z}_n$ which is relatively prime to $n$ has an inverse is the following: Since $\gcd(a,n) = 1$, there exists $x,y \in \mathbb{Z}$ such that $ax + ny = 1$. Now reducing mod $n$ gives that $ax=1$, so the inverse of $a$ is $x$. To find $x$ (and $y$), use the Euclidean Algorithm.