Natural Numbers under Multiplication do not form Group/Proof 2

Proof
that $\struct {\N, \times}$ is a group.

We have that $1 \times 1 = 1$ and so is idempotent.

From Identity is only Idempotent Element in Group it follows that $1$ is the identity of $\struct {\N, \times}$.

Let $x \in \N$ such that $x \ne 0$ and $x \ne 1$.

There exists no $y \in \N$ such that $x \times y = 1$

Hence $\struct {\N, \times}$ does not fulfil.

Hence by Proof by Contradiction $\struct {\N, \times}$ is not a group.