Non-Zero Natural Numbers under Multiplication form Commutative Semigroup

Theorem
Let $\N_{>0}$ be the set of natural numbers without zero, i.e. $\N_{>0} = \N \setminus \left\{{0}\right\}$.

The structure $\left({\N_{>0}, \times}\right)$ forms an infinite commutative semigroup.

Closure
Take the definition of natural numbers $\N$ as a naturally ordered semigroup $\left({\N, +, \le}\right)$.

From Multiplication in Naturally Ordered Semigroup has No Proper Zero Divisors:
 * $\forall m, n \in \N: m \times n = 0 \iff m = 0 \lor n = 0$

It follows that:
 * $\forall m, n \in \N_{>0}: m \times n \ne 0$

and so:
 * $\forall m, n \in \N_{>0}: m \times n \in \N_{>0}$

That is, $\left({\N_{>0}, \times}\right)$ is closed.

Associativity
Natural Number Multiplication is Associative.

Commutativity
Natural Number Multiplication is Commutative.

Infinite
We have that the Natural Numbers are Infinite.

Then we have that Infinite if Injection from Natural Numbers.

The mapping $s: \N \to \N_{>0}: s \left({n}\right) = n + 1$ is such an injection.

Hence $\N_{>0}$ is infinite.