Let $S = \{ n \in {\mathbb N} : n \geq 1 \}\text{.}$ Then $1 \in S\text{.}$ Assume that $n \in S\text{.}$ Since $0 \lt 1\text{,}$ it must be the case that $n = n + 0 \lt n + 1\text{.}$ Therefore, $1 \leq n \lt n + 1\text{.}$ Consequently, if $n \in S\text{,}$ then $n + 1$ must also be in $S\text{,}$ and by the Principle of Mathematical Induction, and $S = \mathbb N\text{.}$
We must show that if $S$ is a nonempty subset of the natural numbers, then $S$ contains a least element. If $S$ contains 1, then the theorem is true by Lemma A.2.5. Assume that if $S$ contains an integer $k$ such that $1 \leq k \leq n\text{,}$ then $S$ contains a least element. We will show that if a set $S$ contains an integer less than or equal to $n + 1\text{,}$ then $S$ has a least element. If $S$ does not contain an integer less than $n+1\text{,}$ then $n+1$ is the smallest integer in $S\text{.}$ Otherwise, since $S$ is nonempty, $S$ must contain an integer less than or equal to $n\text{.}$ In this case, by induction, $S$ contains a least element.