Prove $\Bbb{N}$ is infinite from Peano axioms Let there be a set $\Bbb{N}$ defined by these 3 axioms:
There exists a set $\Bbb{N}$ such that $1\in \Bbb{N}$ and a function $s:\Bbb{N}\rightarrow\Bbb{N}$ satisfying these properties:
$$\not\exists n\in\Bbb{N}:s(n)=1 \tag{i}$$
$$s \text{ is injective} \tag{ii}$$
$$\text{Let  $G\subseteq\Bbb{N}$ be a set. Suppose  $1\in G$.Then if $g\in G\Rightarrow s(g)\in G$ holds, then $G=\Bbb{N}$}\tag{iii}$$
What I am looking for is to prove that $\Bbb{N}$ is in fact infinite. In the book "The Real Numbers and Real Analysis" from E.D.Bloch there is following "explanation" given.
Suppose $\Bbb{N}$ is finite. $\Bbb{N}=\{1,p\}$. 
Let $s:\Bbb{N}\rightarrow\Bbb{N}$ be defined as follows: $s(1)=p,s(p)=p$. We see that in this case our triplet $(\Bbb{N},1,s)$ doesn't satisfy ii).
Let $s:\Bbb{N}\rightarrow\Bbb{N}$ be defined in other way: $s(1)=p, s(p)=1$. In this case, the triplet $(\Bbb{N},1,s)$ doesn't satisfy i).
This "proof" seems very incomplete to me. Can i actually assume without loss of generality that if $\Bbb{N}$ is finite, then it has only 2 elements? It seems clear that if $s$ must be an injection and $\forall n\in\Bbb{N}:s(n)\neq1$. Then, if $\Bbb{N}$ had finite amount of elements, say $n$, we must always have "some next element to point the $s(n)$ to", but I would like a rigorous mathematical proof of that.
 A: The proof of your book shows the method to prove $\mathbb N$ is infinite. With more detail, you should suppose $\mathbb N=\{1,2,3,\dots, n\}$ and get a contradiction simply as follows: by (i) theres is no $m\in\mathbb N$ such that $s(m)=1$. So that, $1\notin\operatorname{im} s \subseteq \mathbb N$. Hence, you have to find an injection from a set with $n$ elemenets into a set with at most $n-1$ elements, which is impossible. The concluison is that if $\mathbb N$ is finite axiom (ii) is violated.
A: He's doing a proof by induction.
That $1 \in \mathbb N$ means $\{1\}\subset \mathbb N$. And $s:\mathbb N\to \mathbb N$ says that $s(1)$ exists.
Axiom a: says $s(1)\ne 1$.  Let's label this as $p_1=s(1)$. So $\{1, p_1\} \subset \mathbb N$ and $\{1\}\subsetneq \mathbb N$..
Now $s(p_1)$ must exist.  But Axiom a: says $s(p_1) \ne 1$ and Axiom b: says $s(p_1) \ne p_1$.  Let's label this as $p_2$.  So $\{1,p_1,p_2\} \subset \mathbb N$ and $\{1,p_1\}\subsetneq \mathbb N$.
Propostion:  For any $n$ in... well, for any number that we can count [!!!] there is $N_k\subsetneq \mathbb N$ where $N_k =\{1,p_1=s(1), p_2=s(p_2), ... p_k=s(p_{k-1})\}$.  And as $N_k\subsetneq \mathbb N$ then $\mathbb N$ must have at least as many elements as $N_k$ which has $k$ elements and therefore $\mathbb N$ can not have any finite number of elements.
Pf:  By axiom b:  $s(p_k)\not \in N_k$.
So that's that.
But.... it's circularly pointless to call this a formal proof.  This axiom declares the natural numbers exists so how can I use them in an induction proof.  
And the word "infinite" hasn't been well defined.
But it does show us that that natural numbers behave with the properties that an infinite set ought to.  And further if we define a set $G$ as "infinite" if there exists an injective $f:\mathbb N \to G$, then that will be a valid definition.  (I think.  We still don't have  a definition for "finite" and if we define "finite" as such an injection does not exist it might be hard to prove that to mean what ... we think it means.  We can do it, but it's not as straightforward as we'd like.)
