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  "body": "\n## Proof\n\nWe argue by [[induction|0ASJVEFA9J]] on $n$. The claim for $n$ is as follows. For every [[field|0QA8CHH8MG]] $K$ and every $f \\in K[x]$ of degree $n$, there is a splitting field $L$ of $f$ over $K$ with $[L : K] \\le n!$.\n\n1. Case $n = 1$. Write $f = c x + d$ with $c \\ne 0$, and let $a_1 = -d/c \\in K$. Then $f = c (x - a_1)$, so $f$ [[splits|0SHMC00RJV]] over $K$.\n2. Since $a_1 \\in K$, the [[generated field|0MA5B744K3]] $K(a_1)$ is $K$. So $L = K$ is a [[splitting field|02NZZ1J0FH]] of $f$, and $[L : K] = 1 = 1!$.\n3. Case $n \\ge 2$. Assume the claim for $n - 1$, and let $f \\in K[x]$ have degree $n$.\n4. [[There is an extension|0VG02M1FEZ]] $K'/K$ with $[K' : K] \\le n$ and a root $a_1 \\in K'$ of $f$. Let $K_1 = K(a_1) \\subseteq K'$.\n5. By the [[tower law|0A7KJ68XHT]], $[K' : K] = [K' : K_1] \\cdot [K_1 : K]$, so $[K_1 : K] \\le [K' : K] \\le n$.\n6. By the [[factor theorem|0K3MWA7DMC]] in $K_1[x]$, $f = (x - a_1) g$ with $g \\in K_1[x]$. By the [[degree of a product|0T233AZ1GE]], $\\deg g = n - 1$.\n7. By the induction hypothesis, $g$ has a splitting field $L$ over $K_1$ with $[L : K_1] \\le (n-1)!$. So $g = c (x - a_2) \\cdots (x - a_n)$ with $c, a_i \\in L$, and $L = K_1(a_2, \\dots, a_n)$.\n8. Then $f = c (x - a_1)(x - a_2) \\cdots (x - a_n)$ in $L[x]$, so $f$ splits over $L$, with roots $a_1, \\dots, a_n$.\n9. Let $F$ be the subfield of $L$ generated by $K$ and $a_1, \\dots, a_n$. We show $F = L$. Here $F \\subseteq L$ because $F$ is a subfield of $L$.\n10. By the [[structure of simple extensions|0ZN39QDTP9]], $K_1 = \\set{ h(a_1) : h \\in K[x] }$. These elements lie in $F$, so $K_1 \\subseteq F$.\n11. So $F$ is a subfield of $L$ that contains $K_1$ and $a_2, \\dots, a_n$. Since $L = K_1(a_2, \\dots, a_n)$ is the smallest such subfield, $L \\subseteq F$.\n12. So $L = K(a_1, \\dots, a_n)$, and with Step 8, $L$ is a splitting field of $f$ over $K$.\n13. By the tower law and Steps 5 and 7, $[L : K] = [L : K_1] \\cdot [K_1 : K] \\le (n-1)! \\cdot n = n!$.\n14. This completes the induction.\n\n## Idea\n\nAdjoin one root of $f$, which costs a factor of at most $n$ in the degree. Over the bigger field, $f$ is $x - a_1$ times a polynomial of degree $n - 1$, and induction handles that polynomial. The degrees multiply by the tower law, giving at most $n \\cdot (n-1)! = n!$.\n",
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