Invertible Elements under Natural Number Multiplication

Theorem
Let $\N$ be the natural numbers.

Let $\times$ denote multiplication.

Then the only invertible element of $\N$ for $\times$ is $1$.

Proof
Suppose $m \in \N$ is invertible for $\times$.

Let $n \in \N: m \times n = 1$.

Then from Natural Numbers have No Proper Zero Divisors:
 * $m \ne 0$ and $n \ne 0$

Thus, $1 \le m$ and $1 \le n$.

If $1 \le m$ then from Ordering on Natural Numbers is Compatible with Multiplication:
 * $1 \le n < m \times n$

This contradicts $m \times n = 1$.

The result follows.