Every integer $n>1$ has a prime divisor; indeed the least divisor of $n$ that exceeds $1$ is prime

Let $n\in \mathbb{Z}$ with $n>1$, and put

$S:=\{d\in \mathbb{Z}:d\mid n\text{ and }d>1\}$

(Divisibility in $\mathbb{Z}$: $d\mid a$ when $a=dq$ for some integer $q$). Then $S$ is nonempty and has a least element $q$, and $q$ is prime (Prime and composite integers: $p$ is prime when $p>1$ and its only positive divisors are $1$ and $p$). In particular every integer greater than $1$ has a prime divisor.

• Well-ordering is used through the integer bridge, not directly. The well-ordering principle gives a least element of a nonempty subset of $\mathbb{N}$, whereas $S$ is a set of integers. A nonempty set of integers bounded above has a greatest element, and a nonempty set of integers bounded below has a least element is the published lemma that crosses that gap, and its Statement covers the lower-bound case explicitly, so it is what this proof cites.

• The bound $q$ is genuinely the least divisor above $1$, not merely some prime divisor. That is what makes the argument work: minimality is the only reason $q$ has no proper divisor above $1$, since any such divisor would divide $n$ as well and be smaller.

• The hypothesis $n>1$ cannot be dropped. At $n=1$ the set $S$ is empty, since $d\mid 1$ forces $d=1$ or $d=-1$ ($(\mathbb{Z},\cdot ,1)$ is a commutative monoid whose group of units is $\{1,-1\}$; equivalently $u\mid 1$ holds exactly for $u=1$ and $u=-1$) and neither exceeds $1$; so $1$ has no prime divisor, which is exactly why the empty product is the right factorisation of $1$ in Every integer $n\ge 1$ is a finite product of primes: there are $r\in \mathbb{N}$ and a list $p:r\to \mathbb{Z}$ of primes with $n={\prod }_{i<r}{p}_i$, the case $n=1$ being the empty product. At $n=0$ every integer divides $0$, so $S$ is the whole set of integers greater than $1$; it does have a least element, namely $2$, but a statement about the divisors of $0$ is of no use below and the lemma is not claimed there.