How to prove that doesnt exist a natural number such that is equal to it successor from Peano axioms? Im getting a hard time trying to prove the general for any natural number $n$ such that
$$\nexists n\in\Bbb N: S(n)= n$$
From the second Peano axiom we know that
$$\nexists n\in\Bbb N: S(1)= n$$
and from the third axiom that
$$S(a)=S(b)\to a=b$$
I really dont have a clear clue about how to prove this. Someone tell me that I must use the "axiom" number 5, the induction. But the induction need the existence of a second natural number more than $1$ (or more than $0$ using the axioms that use it) that hold the implicit difference on the axiom 2. Example:
$$S(1)\ne 1 \to S(1)=2$$
But now, how to prove that $S(2)\ne 2$? Thank you in advance.
 A: The successor map $\sigma : \mathbb{N} \rightarrow \mathbb{N}$ is injective. This is an axiom. In addition, as far as the base case goes, $\sigma(1) \neq 1$; this is also an axiom since $1$ is not the successor of another number. 
Now, we employ induction. Suppose for some $n$, $\sigma(n) \neq {n}$. This implies $$\sigma(\sigma(n)) \neq \sigma(n)$$ since otherwise would contradict the function being injective. 
A: Ok, I solved it using the third axiom. If $S(1)=2$ and $S(2)=2$ then we have that $S(1)=S(2)$ and by the third axiom we have that $1=2$, a contradiction, so $S(2)\ne 2$.
After all it was so simple but I didnt see in first place. What a dumb.
To complete the answer for any number $n$ is enough to see that the natural numbers are defined through the successor function over $1$ (or $0$, depending of the version of the Peano axioms).
So, necesarily, if $S(2)\ne 2$ and $S(1)\ne 1$ then $S(n)\ne n$ by induction.
A: Proof
Indeed, let $X := \{n \in \mathbb{N}: s(n) \neq n\}$
$\Rightarrow 1 \in X$ (by Peano's Axiom 2)
And let $n \in X \Rightarrow  s(n) \neq n$
$\Rightarrow  s(s(n)) \neq s(n)$ (Since s is injective, Peano's Axiom 1)
$\Rightarrow s(n) \in X, \forall n \in X$ 
$\Rightarrow X = \mathbb{N}$ (By Peano's Axiom 3)
