Problem: Let $f(x)$ is a polynomial of order $n$ in $k[x]$ with $k$ is a field and $E$ is the spliting field of $f(x)$. Show that $[E:k] \le n!$
My attempt:
We will prove the problem by induction on $n$. Let $f(x) = ax+b \in k[x], a \ne 0$. Then $E = k(\alpha) = k, \alpha = -b/a \in k$ and $f(\alpha) = 0$, implies that $[E:k] = [k:k] = 1 = 1!$.
Suppose that the problem was proved for every polynomial of order less than $n$. Suppose $f(x) \in k[x]$ with $\deg f(x) = n \ge 2$ and $E = k(\alpha_1,\alpha_2,\dots,\alpha_n)$ is the spliting field of $f(x)$. We have $f(x) = (x - \alpha_1) g(x), g(x) \in k(\alpha_1)[x]$ and $g(\alpha_2) = \dots = g(\alpha_n) = 0$. $E = k(\alpha_1)(\alpha_2,\dots,\alpha_n)$ is the spliting field of $g(x)$, implies that $[k(\alpha_1)(\alpha_2,\dots,\alpha_n)] \le (n-1)!$. Since $f(\alpha_1)=0$ then $[k(\alpha_1):k] \le n$, implies $[k(\alpha_1)(\alpha_2,\dots,\alpha_n)] \le n!$.