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  "body": "\n## Proof\n\n1. Let $k$ be a natural number. For a positive integer $x$, let $D(x)$ be the set of [[positive divisors|0FCWA16FZE]] of $x$. By the definition of the [[divisor function|0YEW47VJ0J]], $\\sigma_k(x) = \\sum_{d \\in D(x)} d^k$.\n2. Since $D(1) = \\set{1}$, we have $\\sigma_k(1) = 1$.\n3. Claim: if $a \\mid bc$ and $\\gcd(a, b) = 1$, then $a \\mid c$. By [[Bézout's identity|0DQVHMYAGA]], $1 = ua + vb$ for integers $u, v$. Then $c = uac + vbc$, and $a$ divides both terms.\n4. Let $m$ and $n$ be [[coprime|0GFKP8F8AG]] positive integers. Define $\\Phi : D(m) \\times D(n) \\to D(mn)$ by $\\Phi(d, e) = de$. This is well defined, since $d \\mid m$ and $e \\mid n$ give $de \\mid mn$.\n5. If $d \\mid m$ and $e \\mid n$, then $\\gcd(d, e) = 1$. Indeed, a common divisor of $d$ and $e$ divides both $m$ and $n$, whose [[greatest common divisor|02WF878BH1]] is $1$.\n6. $\\Phi$ is injective. Suppose $de = d'e'$ with $d, d' \\in D(m)$ and $e, e' \\in D(n)$. Then $d \\mid d'e'$ and $\\gcd(d, e') = 1$, so $d \\mid d'$ by step 3.\n7. In the same way $d' \\mid d$. Both are positive, so $d = d'$, and then $e = e'$.\n8. $\\Phi$ is surjective. Let $c \\in D(mn)$, let $d = \\gcd(c, m)$, and write $c = d c'$ and $m = d m'$. Then $\\gcd(c', m') = 1$, since a common divisor $g \\ge 1$ makes $gd$ a common divisor of $c$ and $m$. Then $gd \\le d$, so $g = 1$.\n9. From $c \\mid mn$ we get $d c' \\mid d m' n$, so $c' \\mid m' n$. By step 3, $c' \\mid n$. So $c = \\Phi(d, c')$ with $d \\in D(m)$ and $c' \\in D(n)$.\n10. So $\\Phi$ is a bijection. Hence $\\sigma_k(mn) = \\sum_{d \\in D(m)} \\sum_{e \\in D(n)} (de)^k = \\left(\\sum_{d \\in D(m)} d^k\\right) \\left(\\sum_{e \\in D(n)} e^k\\right)$.\n11. This is $\\sigma_k(m) \\sigma_k(n)$. With step 2, $\\sigma_k$ is a [[multiplicative function|0AP80TT660]].\n\n## Idea\n\nWhen $m$ and $n$ share no factor, each divisor of $mn$ splits in exactly one way as a divisor of $m$ times a divisor of $n$. So the sum of $k$-th powers over the divisors of $mn$ factors as a product of two sums.\n",
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