I wish to prove that if $x$ is order $\phi(n)$ in $\mathbb{Z}/n \mathbb{Z}$, all its inverses are powers of $x$. Where $\phi(n)$ is the Euler totient function that tells us how many invertible elements there are. Below is what I think is a good proof, but I am still learning and wish to grow:
First of all, there are precisely $\phi(n)< n$ invertible elements in $\mathbb{Z}/n \mathbb{Z}$. We know that the product of two or more invertible elements is again invertible. The condition is that every single element is invertible. Now notice that we know that:
$$x^{\phi(n)} \equiv1 \mod n $$ Since $1$ is its own inverse, it is invertible, the same must then hold for $x^{\phi(n)}$. Since a product of $x$'s leads to an invertible element, we then also know that $x$ must be invertible. Observe that every power of $x$ is invertible, since it consists of a product of invertible elements, let us list them:
$$ 1, x^1, x^2, x^3 ... x^{\phi(n)-1} $$ There are precisely $\phi(n)$ such elements and they are all invertible. However, this is precisely the amount of invertible elements there were in $\mathbb{Z}/n \mathbb{Z}$. We have therefore given a complete list and the full set of invertible elements is simply the list of powers of $x$.