degree of splitting field of p(q(x))

Let $$p(x), q(x) \in F[x]$$ be two polynomials with $$\operatorname{deg}p(x)=m$$ and $$\operatorname{deg}q(x)=n$$. Prove that the splitting field E of $$p(q(x))$$ has a degree that satisfies $$[E:F] \le m!(n!)^m$$

I know that the splitting field $$E$$ of $$p(x)$$ with degree $$n$$ over $$F$$ has property $$[E:F] \le n!$$

And I don't learn Galois theory. So I want to solve the problem only with the definition of splitting field and field extension. Help me!

I hope you agree that the splitting field $$K$$ of $$p(x)$$ has degree $$\le m!$$ over $$F$$.
If $$p(q(\alpha))=0$$ then $$q(\alpha)=\beta_i$$ for some $$i$$ where $$\beta_1,\ldots, \beta_m$$ are the roots of $$p(x)=0$$ in $$K$$. So $$\alpha$$ is a zero of some $$h_i(x)=q(x)-\beta_i$$. Let $$K_1$$ be the splitting field of $$q(x)-\beta_1$$ over $$K=K_0$$, let $$K_2$$ be the splitting field of $$q(x)-\beta_2$$ over $$K_1$$, etc. Then $$|K_{i+1}:K_i|\le n!$$ for each $$i$$, and $$\alpha\in K_m$$. So $$E\subset K_m$$ (indeed $$E=K_m$$).
• Thank you for help. But I have a question. Where can I find the m!? I understand that $[K_m:K_0] \le (n!)^m$ I can't find m! – Pearl Oct 3 '18 at 6:07
• @Pearl As I said, $|K:F|\le m!$. – Lord Shark the Unknown Oct 3 '18 at 6:21