Proof. sketch [1503A]

Clearly, $f(x)$ splits completely in $F(\alpha_{1}, \dots, \alpha_{n})$.

Claim: $F \subseteq E \subseteq K$ and $f(x)$ splits completely over $E$ $\implies$ $F(\alpha_{1}, \dots, \alpha_{n}) \subseteq E$.

Proof of the Claim: Suppose $f(x) = b(x - \beta_{1})\cdots(x - \beta_{n})$ with $b, \beta_{1}, \dots, \beta_{n} \in E$. Then we have $$ a(x - \alpha_{1}) \cdots (x - \alpha_{n}) = b(x - \beta_{1}) \cdots (x - \beta_{n}) $$ in $K[x]$. Since $K[x]$ is a UFD, after reordering we have $\alpha_{i} = \beta_{i}$ for all $1 \leq i \leq n$. And $F(\beta_{1}, \dots, \beta_{n}) \subseteq E$. Hence $F(\alpha_{1}, \dots, \alpha_{n}) \subseteq E$.

1. Suppose $F \subseteq E \subseteq F(\alpha_{1}, \dots, \alpha_{n})$ and $f(x)$ splits completely over $E$, then by our claim, $E = F(\alpha_{1}, \dots, \alpha_{n})$. So $F(\alpha_{1}, \dots, \alpha_{n})$ is a splitting field of $f(x)$.
2. Since $f(x)$ splits completely over $E$, by our claim, $F(\alpha_{1}, \dots, \alpha_{n}) \subseteq E$. Moreover, $E$ is a splitting field of $f(x)$, then by definition, $E = F(\alpha_{1}, \dots, \alpha_{n})$.