Proving the Well-Ordering Principle for Natural Numbers I know the WOP is treated like axiom of a natural number, but I was curious if I can prove WOP defined for the set of natural numbers N by following:
Suppose A is a subset of N, which then obeys all axioms from Peano postulates, and let us assume that A does not have a least element.  Then, 1 in A must be a successor to another natural number, which is 0.  However, 0 is neither in A nor N.  (If 0 is in N, then we argue that 0 must be a successor to -1, which is not part of N).  Also, 1 in A being a successor contradicts with the Peano axiom (1 cannot be a successor).  Hence, our assumption that A having no least element is not true.  Therefore, A has a least element.  Without out loss o generality, every subset of the N has a least element.
I understand this is silly proof, but I wanted to make sure that I have reasoning for WOP in N before moving on.

In my opinion, you have essentially restated the axiom by making the same assumption in a more subtle fashion (assumptions about the naturals). Generally, axioms are axioms because they cannot be proven independently using the other axioms. For instance, if you are familiar with analysis, the existence of suprema in the reals is typically axiomatic, but the existence of infima is not, since it can be proven from the suprema axiom.
