I think I've read enough about the Peano axioms and I also checked the similar posts here on StackExchange, but all I want to know is why induction can't be validated for all n by writing down the finite logical chains. We don't really need the well-ordering principle...
If we assume to the contrary that there exists an m st. P(m) is false, we may reach the clear contradiction by writing out the corresponding logical chain P(1)->P(2)->...P(m), and so P(m) is true.
What's wrong with this?
Why should there in fact exist such a chain starting from $1$ and reaching all the way up to $m$? This uses an assumption about $\mathbb{N}$, which is ... exactly the well-ordering principle (or something equivalent to it).
This can feel very slippery at first. I think the best way to clarify the situation is to look at a structure which is like the natural numbers except for induction. Since induction fails in such a structure, there must be something special about the natural numbers which lets us prove that induction holds of them.
Specifically, I want to look at a discrete ordered semiring with a simple failure of induction. Here's an example I quite like:
Consider the set $\mathfrak{X}$ of polynomials in a single variable $t$ with integer coefficients, which are either zero or have positive leading term.
There are obvious notions of addition and multiplication of elements of $\mathfrak{X}$ (just the usual operations on polynomials). Moreover, $\mathfrak{X}$ carries a natural ordering: we set $p<q$ iff $\lim_{a\rightarrow \infty}p(a)<\lim_{a\rightarrow\infty}q(a)$. (It's a good exercise to give an equivalent definition of this ordering in terms of coefficients alone.) This turns $\mathfrak{X}$ into a discrete ordered semiring, just like $\mathbb{N}$.
However, induction fails in $\mathfrak{X}$: consider for example the property
$P(a)\equiv$ "For every $b<a$ there is some $c$ such that either $c+c=b$ or $c+c+1=b$."
That is, "Every number $<a$ is either even or odd." In any discrete ordered semiring whatsoever this is trivially satisfied by $0$ and satisfies $\forall x(P(x)\rightarrow P(x+1))$. However, $\forall x(P(x))$ is not true in $\mathfrak{X}$: consider the monomial $t$.
So there is something special about $\mathbb{N}$ which doesn't come from its discrete ordered semiring structure alone but needs to be considered separately, and this is exactly captured by the well-ordering principle. Put another way:
Since induction holds in $\mathbb{N}$ but fails in $\mathfrak{X}$, any argument that induction holds in $\mathbb{N}$ must use some property of $\mathbb{N}$ which doesn't hold of $\mathfrak{X}$. This is what the well-ordering principle provides us.
