Suppose $L/K$ is a finite extension. $G$ is a finite group of $K$-automorphisms of $L$. Denote by $L^G$ the field elements of $L$ fixed by action of $G$. For any $\alpha \in L$ we write $f(t, \alpha) = \prod_{g \in G} (t-g(\alpha))$.
Suppose $L = K(\alpha)$. I want to prove that $L^G$ is generated over $K$ by the coefficients of $f$.
Here are some of my thoughts.
Note that the degree of this polynomial is $\# G$. Let $K(c_1,...,c_{\#G})$ be the field $K$ adjoin the coefficients of the polynomial. All of those coefficients lie inside $L^G$ (as the polynomial lies in $L^G[X]$ by construction), meaning that $K(c_1,...,c_{\#G}) \subseteq L^G$. We know $[L : L^G] = \# G$ from Artin's theorem.
I want to now either:
Establish the reverse inclusion, and show $K(c_1,...,c_{\#G}) \supseteq L^G$, which will give me $K(c_1,...,c_{\#G}) = L^G$.
Show that the degree $[L^G : K(c_1,...,c_{\# G})$ is $1$, which will also give me $K(c_1,...,c_{\#G}) = L^G$.
I am unsure how to proceed down either of these routes and would appreciate some guidance.
