Proof. sketch [1507A]

Let $F$ be the base field and $f(x) \in F[x]$ be the polynomial with $\deg f(x) = n$. Let $\alpha_{1}, \dots, \alpha_{n}$ be the roots of $f(x)$ and $E = F(\alpha_{1}, \dots, \alpha_{n})$ be the splitting field of $f(x)$.

Then $[F(\alpha_{1}) : F]$ has degree at most $n$ since $1, \alpha_{1}, \dots, \alpha_{1}^{n}$ is not linearly independent ($f(\alpha_{1}) = 0$). Moreover, by Homework 2, $[F(\alpha_{1}, \alpha_{2}) : F(\alpha_{1})] < [F(\alpha_{1}) : F] \leq n \implies [F(\alpha_{1}, \alpha_{2}): F(\alpha_{1})] \leq n-1$. If we keep doing so, we will finally arrive at $$ \begin{align*} [E : F] &= [F(\alpha_{1}, \dots, \alpha_{n}) : F(\alpha_{1}, \dots, \alpha_{n-1})] \cdots [F(\alpha_{1}, \alpha_{2}) : F(\alpha_{1})][F(\alpha_{1}) : F] \\ &\leq 1 \cdots \cdot (n-1)n \\ &= n! \end{align*} $$