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  "body": "\n## Proof\n\nWe argue by [[strong induction|0KMVFSHFGZ]] on $n$.\n\n1. For a natural number $n$, let $P(n)$ say: if $n \\ge 1$, then $n = p_1 p_2 \\cdots p_k$ for some [[prime numbers|0SN0MAEKYX]] $p_1, \\dots, p_k$ with $k \\ge 0$.\n2. Let $n$ be a natural number, and assume that $P(m)$ holds for every natural number $m \u003c n$. We prove $P(n)$.\n3. If $n = 0$, then $P(n)$ holds, because its hypothesis $n \\ge 1$ is false.\n4. If $n = 1$, take $k = 0$. The empty product is $1$, so $n$ is a product of primes.\n5. Let $n \\ge 2$. By the [[existence of a prime divisor|0SPBEDVNW1]], there is a prime $p$ with $p \\mid n$. Write $n = p m$ with an integer $m$.\n6. Since $n \u003e 0$ and $p \u003e 0$, $m \\ge 1$. Since $p \\ge 2$, $n = p m \\ge 2m \u003e m$. So $1 \\le m \u003c n$.\n7. By the induction hypothesis $P(m)$, $m = q_1 \\cdots q_j$ for some primes $q_1, \\dots, q_j$ with $j \\ge 0$.\n8. Then $n = p q_1 \\cdots q_j$ is a product of $j + 1$ primes, so $P(n)$ holds.\n9. By strong induction, $P(n)$ holds for all $n$. So each positive integer is a product of primes.\n10. Listing these primes in increasing order gives $p_1 \\le \\dots \\le p_k$ with the same product. This is a [[prime factorization|0AEKDFCYFA]] of $n$.\n\n## Idea\n\nSplit off one prime factor of $n$. What is left is a smaller positive integer, which is a product of primes by induction. Putting the prime back gives a product of primes for $n$.\n",
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