"Man created the natural numbers, and all other kind of stuff."
Shaito san (as noted by George on the fourth moon of the Year of the Shark).

Well-Ordering Principle and Least Element
Set Theory has spread its tentacles to every corner of Mathematics, even in the venerable Number Theory which is (almost) as old as Mathematics itself. See how Burton ("Elementary Number Theory", 2002) formulates an ancient theorem:

"Theorem 1.1 Archimedean property. If a and b are any positive integers, then  there exists a positive integer n such that na >= b.
Proof. Assume that the statement of the theorem is not true, so that for some a and b, na < b for every positive integer n. Then the set 
S = {b — na | n a positive integer} 
consists entirely of positive integers. By the Well-Ordering Principle, S will possess a least element, say, b — ma. Notice that b — (m + 1)a also lies in S, because S contains all integers of this form. Furthermore, we have 
b — (m + l)a = (b — ma) — a < b — ma 
contrary to the choice of b — ma as the smallest integer in S. This contradiction arose out of our original assumption that the Archimedean property did not hold; hence, this property is proven true."

The theorem is in itself crystal clear. Or so it seems.
If you take two distinct numbers (a and b), and multiply one of these numbers (a) by a third number (c), then the result will be, at least in some cases, either equal or greater than the other number (b). 
Maybe not so clear after all. It sounds like the theorem is saying that the result could be anything! Which is certainly true. It can be less, equal, or greater than b. What kind of theorem is that?!
Look now at the proof: "there exists a positive integer" has suddenly become "for every positive integer"! That is cheating! The theorem never said that the result would always be equal or greater than b, only that it might!

Okay, even if that was the fifth edition of the book (!), maybe Burton somehow overlooked this error, and meant something more like:
Theorem: If a and b are any positive integers, then any positive integer n is such that na >= b."
I still would not know why anyone would want such a theorem, but then, maybe that is why I am not a mathematician.
Let us look now at the proof once again.
Notice the Well-Ordering Principle that stands prominently in the proof. It tells us that any well-ordered set has a least element. How do we know that S is a well-ordered set? Simple, the definition of the principle tells us very clearly: "Every nonempty set S of nonnegative integers contains a least element". Since the elements of S are all positive integers, the case is closed.
But then I start to wonder: what does na mean? Multiplication of a by n, right? so, by definition, na will always be greater than a. We have therefore the following possibilities:
a equal or greater than b: in both cases na will be greater than b, and we would have proved the (new) theorem.
a smaller than b: but how much smaller? Even then, na can turn out to be greater than b. In which case, we would also have proved the theorem.
Please notice that we did not need the Principle of Well ordering (and the existence of a least number). Just like Archimedes also never did. The "nature" of natural numbers was in itself sufficient to bear the weight of the proof.
This clearly shows that, at least in some cases, the principle is completely superfluous. It expresses a property which is inherent to natural numbers in that they would not be natural numbers if they did not possess it. In other words, all the principle says is that natural numbers are... natural numbers.

The idea of a least number sounds very powerful, until we realize that it does not in fact add anything to our understanding of the situation. Dealing with natural numbers, we indeed always know that there will be a least number. Always. In any situation and in any set of natural numbers. If we could use that fact to prove what we want, we would be able to prove anything concerning natural numbers.
Because it is always true it does not tell us anything about the rest of the proof. Even if the proof is invalid, it will still be true that a set of natural numbers has a least element.
The idea of a least element is therefore useless, and in that, it resembles suspiciously the idea of a first element as analyzed above in the entry about von Neumann.
Let us look now at the following point.
"Theorem 1.2 First Principle of Finite Induction. Let S be a set of positive integers with the following properties: 
(a) The integer 1 belongs to S. 
(b) Whenever the integer k is in S, the next integer k + 1 must also be in S. 
Then S is the set of all positive integers."
Here is the proof:
"Let T be the set of all positive integers not in S, and assume that T is nonempty. The Well-Ordering Principle tells us that T possesses a least element, which we denote by a. Because 1 is in S, certainly a > 1, and so 0 < a — 1 < a. The choice of a as the smallest positive integer in T implies that a — 1 is not a member of T, or equivalently that a — 1 belongs to S. By hypothesis, S must also contain (a — 1) + 1 = a, which contradicts the fact that a lies in T. We conclude that the set T is empty and in consequence that S contains all the positive integers."

This is a perfect example of selective thinking. 
k will always be greater than 1, and k+1 greater than k. What else could you possibly need to say that nothing smaller than k could be anything else but 1. That was your assumption to start with. The concept of least element cannot be used as part of the proof. because it is itself the consequence of it.
If any k > 1, than nothing can be smaller than k.
And if it is possible, then there is no reason that some positive numbers smaller than k, could not be elements of T.