Abstract Algebra (splitting fields) Suppose $f(x)∈F[x]$ is a polynomial of degree $n$. Let $k$ be its splitting field. Prove that $[K:F]\leq n!$. 
I figure I am supposed to use induction on n, but I am not sure how to go about doing that. Thanks for any help.
 A: Hints:
Assume $\;f(x)\in\Bbb F[x]\;$ is irreducible, $\;\deg f=n\;$:
$$\text{Our polynomial's a root in the field}\;\;\Bbb F_1:=\Bbb F[x]/\langle f(x)\rangle\;,\;\;\text{namely}\;\;\alpha:=\overline x:=x+\langle f(x)\rangle$$
We can write $\;f(x)=(x-\alpha)q_1(x)\in\Bbb F_1[x]\;\text{ and}\;\;\deg q_1=n-1$
Note that $\;[\Bbb F_1:\Bbb F]=n\;$ . Assume $\;q_1(x)\;$ is irreducible over $\;\Bbb F_1\;$:
$$\text{Our polynomial's a root in the field}\;\;\Bbb F_2:=\Bbb F_1[x]/\langle q_1(x)\rangle\;,\;\;\text{namely}\;\;\beta:=\overline x:=x+\langle q_1(x)\rangle$$
Note, again, that $\;[\Bbb F_2:\Bbb F_1]=n-1\implies [\Bbb F_2:\Bbb F]=n(n-1)\;$
And etc.
Note that after the first step, if $\;q_1(x)\;$ happens to be reducible over $\;\Bbb F_1\;$ then, taking one of its irreducible factors, we get an extension of degree less than $\;n(n-1)\;$ (Fill in details)
A: If $\alpha$ is a root of $f$ which is not in $F,$ then $[F(\alpha): F] = n.$ For any other root $\beta,$ its minimal polynomial over $F[\alpha]$ has degree at most $n-1$. Applying induction properly should yield an upper bound for your solution.
A: You might remember the following
Lemma. Let $f(X) \in F[X]$ ($F$ here is a field) a polynomial of degree $n \ge 1$. Then there exists a finite field extension $E$ of $F$ in which $f(X)$ has a root. Moreover it is $[E:F] \le n$. 
and then conclude using induction on $n$.
