discrete mathematics - Prove if a (mod n) has a multiplicative inverse, then it's unique

Assume that an integer $a$ has a multiplicative inverse modulo an integer $n$. Prove that this inverse is unique modulo $n$.

I was given a hint that proving this Lemma:
\begin{align}
n \mid ab \ \wedge \ \operatorname{gcd}\left(n,a\right) = 1
\qquad \Longrightarrow \qquad
n \mid b
\end{align}
should help me in finding the answer.

Here are my steps in trying to solve the problem:
\begin{align}
\operatorname{gcd}\left(n,a\right) = 1
& \qquad \Longrightarrow \qquad
sn + ta = 1
\qquad \Longrightarrow \qquad
sn = 1 - ta \\
& \qquad \Longrightarrow \qquad
1 \equiv ta \mod n

\qquad \Longrightarrow \qquad
ta \equiv 1 \mod n .
\end{align}
I know that having the GCD of m and a equal to 1 proves there is a multiplicative inverse mod n, but I'm not sure on how to prove $n \mid b$ with and how it helps prove the multiplicative inverse is unique.

Answer

For the first part note as $(n,a) = 1$, then there exist $x,y \in \mathbb{Z}$, s.t. $nx + ay = 1$. Multiply both sides by $b$ and you will get $nxb + (ab)y = b$. The LHS is obviously divisible by $n$, so then the RHS must be too. Hence $n \mid b$.

This thing proves that the inverse exist. To note that it's unique modulo $n$ assume that $aa_1 \equiv 1 \pmod n$ and $aa_2 \equiv 1 \pmod n$. Then we have that there exist integer $x,y$ s.t. $aa_1 + nx = 1$ and $aa_2 + ny = 1$. Subtracting the two equations we have that $a(a_1 - a_2) = n(y-x) \implies a(a_1 - a_2) \equiv 0 \pmod n \implies a_1 \equiv a_2 \pmod n$. Hence the multiplicative inverse is unique in $\mathbb{Z}_n$