Induction and well ordering principle

Question

Induction stays that is we have $P(0)$ is true and $\forall n \in \mathbb{N},\ P(n) \implies P(n+1)$ then $P(n)$ is true for every $n \in \mathbb{N}$.

I found a proof that Induction conclusion is true if we suppose that the well ordering principle is true. And the proof goes like this : Suppose that the induction conclusion is false. Meaning that the set $\{ n,\ P(n) \text{ is false } \}$ is non empty. By the well ordering principle, it contains a least element $n_0$ . $n_0 >0$ because $P(0)$ is true. Then $P(n_0 - 1)$ must be true otherwise it contradicts the definition of $n_0$ but using the second point of the induction, we must also have $P(n_0)$ true, hence the contradiction.

I have three questions :

1. Is this proof correct (there is no hidden circularity in the reasoning)
2. Is there an implication the other way around ?
3. Which principle can be first implied from Peano axioms ? Induction, or Well ordering principle ?

Thank you for your answer

Answer 1

The only minor gap if at all is that you assume that every natural number $\ne0$ has a predecessor. In fact this property is what distinguishes the naturals from other well-orders infinite sets.

Answer 2

1. It's correct, you've basically proved principle of weak induction assuming the well ordering principle.
2. You can prove the well ordering principle assuming the principle of weak induction. As the induction step, you are given that all non-empty subsets of $\mathbb{N}$ of size $k - 1$ have a least element, and in this step you must show that all non-empty subsets of $\mathbb{N}$ of size $k$ have a least element. This is trivial to show.
3. None of those two are implied by Peano's axioms. These are axioms (although equivalent) on their own.