Identity Element of Natural Number Multiplication is One

Theorem
The identity of multiplication of natural numbers is $1\equiv S(0)$:
 * $\exists 1 \in \N: \forall n \in \N: n \times 1 = n = 1 \times n$

Proof
First, consider the element $1 \equiv S(0)$, where $S:\N \to \N \setminus \{0\}$ is the successorship mapping.

By Definition of Multiplication of Natural Numbers:
 * $\forall n\in \N: n \times 1 = n + (n\times 0)=n$

so $1$ is a left identity of $\times$ in $\N$.

Let $\Phi$ be the unary predicate defined by $\Phi(k) \iff (1\times k=k)$.

Clearly, $\Phi(0)$ holds, because $1\times 0 = 0$ by the definition of multiplication.

Now presume $\Phi(k)$ holds for some $k\in \N$.

It follows that:
 * $k = 1 \times k \implies S(k) = S(1\times k) = 1+(1\times k)$

By the definition of multiplication, this is equal to $1\times S(k)$, which finally implies that $\Phi(S(k))$ holds.

So, by the Principle of Mathematical Induction $1$ is an identity of $\times$ in $\N$.