Proof
(related to Proposition: Algebraic Structure Of Natural Numbers Together With Multiplication)
The proof will consist of four steps:
This follows from the corresponding proposition.
This follows from the corresponding proposition.
Step 3: Demonstrate that \( ( \mathbb N\setminus\{0\}, \cdot )\) forms a commutative monoid.
 Step 1 shows that "\( \cdot \)" is associative, so \( ( \mathbb N\setminus\{0\}, \cdot )\) forms a
semigroup.
 Following the definition of multiplication, \(n \cdot 1=1\cdot n=n\).
 Thus, the element \(1\in\mathbb N\setminus\{0\}\) is the identity element of this semigroup.
 Together with step 2, this shows that \( ( \mathbb N\setminus\{0\}, \cdot )\) forms a commutative monoid.
Step 4: Show that \( ( \mathbb N\setminus\{0\}, \cdot )\) is cancellative.
This follows from the corresponding proposition.
These four steps show altogether that the set \( ( \mathbb N\setminus\{0\}, \cdot )\),
i.e. the set of natural numbers without the element \(0\), together with the multiplication
"\( \cdot \)" as a binary operation, forms a cancellative commutative monoid.
∎
Thank you to the contributors under CC BYSA 4.0!
 Github:

References
Bibliography
 Kramer Jürg, von Pippich, AnnaMaria: "Von den natürlichen Zahlen zu den Quaternionen", SpringerSpektrum, 2013