Case 1: $F$ is infinite. Write $K = F(\beta, \gamma)$ (the general case reduces to this by induction). We find $\alpha = \beta + c\gamma$ for some $c \in F$.

Let $f, g$ be the minimal polynomials of $\beta, \gamma$ over $F$, with roots $\beta = \beta_1, \ldots, \beta_m$ and $\gamma = \gamma_1, \ldots, \gamma_n$ in $\overline{F}$ (all distinct by separability).

For $i = 1, \ldots, m$ and $j = 2, \ldots, n$, the equation $\beta_i + c\gamma_j = \beta + c\gamma$ gives $c = (\beta - \beta_i)/(\gamma_j - \gamma)$. Since $F$ is infinite, we can choose $c \in F$ avoiding all these finitely many values.

Set $\alpha = \beta + c\gamma$. We show $K = F(\alpha)$. Consider $h(x) = f(\alpha - cx) \in F(\alpha)[x]$. Then $h(\gamma) = f(\beta) = 0$, so $\gamma$ is a common root of $h(x)$ and $g(x)$ in $\overline{F}$. By choice of $c$, $\gamma$ is the only common root (if $h(\gamma_j) = 0$ for $j \neq 1$, then $\alpha - c\gamma_j = \beta_i$ for some $i$, i.e., $\beta + c\gamma - c\gamma_j = \beta_i$, giving $c = (\beta - \beta_i)/(\gamma_j - \gamma)$, excluded). So $\gcd(h, g) = (x - \gamma)$ in $\overline{F}[x]$, which means $\gcd(h, g) \in F(\alpha)[x]$ has $\gamma$ as a root. Thus $\gamma \in F(\alpha)$, and $\beta = \alpha - c\gamma \in F(\alpha)$.

Case 2: $F$ is finite. Then $K$ is also finite, and $K^\times$ is cyclic (finite multiplicative subgroup of a field). Any generator $\alpha$ of $K^\times$ satisfies $K = F(\alpha)$. $\blacksquare$