Question is in the title, here's my proof attempt:
Let $n \in \mathbb{N} \setminus \{1\}$ be arbitrary but fixed. Define the following set:
$$S = \{m \in \mathbb{N}: nm \neq 1\}$$
Let $m = 1$. Since $n \in \mathbb{N} \setminus \{1\}$, it follows that $n \neq 1$. So, $n \cdot 1 = n \neq 1$. Hence, $1 \in S$.
Let $m \in S$. Then, $nm \neq 1$. Suppose that $nm' = 1$, where $m'$ is the successor of $m$. Then:
$$1 = nm' = nm + n$$
Since $n \neq 1$, it follows that $n = l'$ for some $l \in \mathbb{N}$. So:
$$1 = nm+n = nm+l' = (nm+l)'$$
This is a contradiction because it implies that the successor of $nm+l$ is equal to 1 and, by the Peano Axioms, $1$ is not the successor of any natural number. It follows that $nm' \neq 1$. Hence, $m' \in S$.
By the Axiom of Induction, it follows that $S = \mathbb{N}$ and that proves the desired result.
Does the proof above work? If it doesn't, why? How can I fix it?