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  "body": "\n## Proof\n\nWe show that the least divisor of $n$ greater than $1$ is prime.\n\n1. Let $n$ be a natural number with $n \\ge 2$.\n2. Let $D$ be the set of natural numbers $d \\ge 2$ with $d \\mid n$, in the sense of [[divisibility|096PW7Q014]]. Since $n \\mid n$ and $n \\ge 2$, $n \\in D$.\n3. By the [[well-ordering principle|01BX15HSRR]], $D$ has a least element $p$. Then $p \\ge 2$ and $p \\mid n$.\n4. Let $e$ be a positive divisor of $p$ with $e \\neq 1$. Write $p = e f$ with an integer $f$. Since $p \u003e 0$ and $e \u003e 0$, $f \\ge 1$, so $e \\le p$.\n5. Since $e \\neq 1$ and $e \\ge 1$, we have $e \\ge 2$.\n6. Write $n = p g$ with an integer $g$. Then $n = e (f g)$, so $e \\mid n$. With step 5, this gives $e \\in D$.\n7. Since $p$ is the least element of $D$, $e \\ge p$. With step 4, $e = p$.\n8. So the only positive divisors of $p$ are $1$ and $p$, and $p \u003e 1$. Hence $p$ is a [[prime number|0SN0MAEKYX]], and $p \\mid n$.\n\n## Idea\n\nLook at the smallest divisor of $n$ that is larger than $1$. Any divisor of it that is larger than $1$ also divides $n$, so it cannot be smaller. Hence this smallest divisor has no proper divisors, and it is prime.\n",
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