How can I prove that the inverse of $n-1$ in $U(n) = \mathbb{Z}_n^{\times}$ is $n-1$?

Question

Where $U(n)$ is multiplicative group $mod(n)$.

It seems obvious but how can I actually prove it?

From modular arithmetics we have: $(n-1)a = nk+1$, so $a=(nk+1)/(n-1)$, which should be an integer for some integer $k$. How can I get rid of $k$?

Answer 1

To show the inverse of $a$ is $b$, all you need to show is that $ab=1$ in $U(n)$.

So all you need to do is show that $(n-1)(n-1)\equiv 1$ (mod $n$).