Extension With an Element Has Different Images Under Automorphism Generates the Extension

Question

Let $K/F$ be field extension with $[K:F]=n$. Let $a\in K$ be element such that there exist $\sigma_1,\dots,\sigma_n\in \text{Aut}(K/F)$ with $\sigma_i(a)\neq \sigma_j(a)$ whenever $i\neq j$. Here there is no assumption about the extension. I want to show $K=F(a)$. My idea is to consider $f(x)=\displaystyle \prod_{i=1}^n(x-\sigma_i(a))$. Then $f$ divides minimal polynomial of $a$ in $F[x]$. But since minimal polynomial of $a$ has degree at most $n$, then $f$ is minimal polynomial of $a$ and then $K=F(a)$. Is this approach correct?