Proof that $1$ is the least natural number using the Principle of Well Ordering I'm doing a question from a PDF on Abstract Algebra.(The pdf can be found here http://abstract.pugetsound.edu/ 2012 edition). I have to show that the Principle of Well-Ordering implies that $1$ is the smallest natural number; this is only part of the whole question.(Question 14 of Chapter 2).

This is what I got so far. 

Proof.

Let $S$ be an arbitrary subset of the Natural Number excluding one. By the Principle of Well-Ordering there exist a least element $X \in S.$
Now add $1$ to $S.$ $1$ is now the least element. Since there is no $X &lt; 1$ in any such set $S,$ $1$ is the smallest element of any set $S$ and therefore the smallest natural number.
By the Well-Ordering principle, there is a least element $a$ in $\mathbb{N}$. Suppose, to the contrary, that $a&lt;1$. Multiplying both sides of the inequality by $a$, we have $a^2&lt;a$. Since $\mathbb{N}$ is closed under multiplication, $a^2\in\mathbb{N}$, which contradicts the assumption that $a$ is the least element in $\mathbb{N}$.
let S=N{1},by well-ordering principle, S has a least element, namely x. We claim that x>1. Suppose x&lt;1, and x is a natural number, then we have 1-1=0 is also a natural number, contradiction arises. Hence 1