Axiomatically, set $\mathbb{N}$ is constructed via injective function $s:\mathbb{N}\rightarrow \mathbb{N}$ and an element $1\in\mathbb{N}$ and we have that $\forall n\in\mathbb{N}:s(n)\neq1$. Together with this, we are given the axiom of mathematical induction, that is:
Axiom (PMI): Let $S\subseteq\mathbb{N}$ satisfy following properties: $$1\in S\tag{i}$$ $$n\in S\Rightarrow s(n)\in S\tag{ii}$$ then $S=\mathbb{N}$.
Now, in order to talk about well ordering, we should define the ordering we wish to prove is well on $\mathbb{N}$. We define addition:
Definition (addition): $+\mathbb{N}^2\rightarrow \mathbb{N}$ $$a+1=s(a)\tag{iii}$$ $$a+s(b)=s(a+b)\tag{iv}$$
Then, we define strict order on $\mathbb{N}$ by:
Definition (ordering): $$a<b\Leftrightarrow \exists x\in\mathbb{N}:a+x=b\tag{v}$$ $$a\leq b\Leftrightarrow a<b\vee a=b\tag{vi}$$
It is straightforward to show that $\leq$ is total ordering. Now I wish to prove that $\mathbb{N}$ is well ordered by $\leq$. So, I proceed by showing that PCI, also known as Principle of Complete Induction, holds.
Theorem (PCI): Let $S\subseteq \mathbb{N}$ satisfy following properties: $$1\in S\tag{vii}$$ $$\{1\ldots n\}\subseteq \mathbb{N}\Rightarrow s(n)\in S\tag{viii}$$ then $S=\mathbb{N}$.
Proof: Proceed by induction. Because $1\in S$ it follows that $\{1\}\subseteq S$, so the base step holds. Now assume that $\{1\ldots n\}\subseteq S$. By (viii) $s(n)\in S$, thus $\{1\ldots,n,s(n)\}\subseteq S$. By PMI $S=\mathbb{N}$.
So far so good, now I show that this implies the WOP also known as Well-ordering principle.
Theorem(WOP): $\mathbb{N}$ is well ordered by $\leq$, equivallently, any nonempty subset of $\mathbb{N}$ has minimal element with respect to $\leq$.
Proof: Assume there exists some $S\subseteq \mathbb{N}$ such that $S$ has no minimal element. We show that $S$ is empty. Let $P(n)$ be predicate: $$P(n):n\notin S$$ and proceed by PCI. $P(1)$ holds, because (It is straightforward to show by induction) $\forall a\in\mathbb{N}:1\leq a$ and if $1\in S$ then $1$ is minimal element of $S$ which contradicts our assumption. So base step holds. Assume $\forall k\in \{1\ldots n\}:P(k)$. We wish to show that $P(s(n))$ holds. If $P(s(n))$ didn't hold, we would have that $s(n)\in S$ but in that case $s(n)$ is the smallest element in $S$, which is impossible. So by PCI it follows, that $\forall n\in\mathbb{N}:(n)$. But this shows that $S$ is empty. Thus any nonempty subset of $\mathbb{N}$ must have minimal element.
Now, this looks very pretty, but as I red through this few times, i couldn't stop thinking about the set $\{1\ldots n\}$. In fact, what do we mean by set $\{1 \ldots n\}$? This is my "explanation": In fact $$\{ 1\ldots n\}=\{1,s(1),s(s(1)),\ldots,s(s(s(\ldots s(1))))\}$$ So $n$ is some finite composition of $s$ applied to $1$. Trivially $\forall n\in\mathbb{N}:n<s(n)$. Denote $M=\{1\ldots n\}$. Surely there exist some (unique) $n_1\in M$ such that $s(n_1)=n$, next, there is some $n_2\in M$ such that $n_1=s(n_2)$. By this process, we construct finite decreasing sequence $\{n_k\}$ ending at $1$ and we say that $\forall k\in M:k<s(n)$ by transitive property of the order. Most importantly, we are not able to construct any strict lesser number than $n$ itself by applying $s$ to it (because all these numbers are already in $M$) and thus (as said in the inductive step): $\forall k\in \mathbb{N}\setminus M:s(n)\leq k$ and thus $s(n)$ would be the least element in $S$.
If someone was thinking about the term finite, i can definite infiniteness as follows: A set $A$ is infinite if there is a bijection between its proper subset and the set itself. Then finite set is a set which is no infinite. But I am still unable to talk about numbers or how many times do i apply $s$. I could say that we apply $s$ to 1 $n$ times, but again, what is $n$? ... aand we run in circles.
Now, I am really curious about someone's opinion on that, because I see a very biig circular reasoning here. In fact, can i somehow rigorously show that $s(n)$ would be the least element in $S$ in the inductive step of the proof of WOP? Or am i actually completely wrong in something? Honestly, this argument seems weak to me and wouldn't persuade me. Thanks for any useful tips!
Recursively define $K(n) = \{1,2,.. n\}$
$= \{ 1, s(1), s(s(1)),.. s^{n-1}(1)\}. $
$K(1) = \{ 1 \}; K(s(n)) = K(n) \cup \{s(n)\}.$
Similarly, for any other $n$ applications of $s.$