$F \le E$ extension. every element $\alpha \in E - \overline{F}_E$ over $\overline{F}_E$ transcendental

Let $F \le E$ be a field extension. Prove that every element $\alpha \in E \setminus \overline{F}_E$ over $\overline{F}_E$ transcendental.

Note: $\overline{F}_E = \{x \in E : x \text{ is algebraic over } F \}$.

Edit: If $\alpha$ algebraic over $\overline{F}_E$ then $\overline{F}_E (\alpha)$ is algebraic over $\overline{F}_E$ and so $\overline{F}_E$ is algebraic over $F$. Hence $\overline{F}_E(\alpha)$ is algebraic over F and $\alpha$ is algebraic over $F$.

Can you help me?

Suppose that $\alpha$ is algebraic over $\overline F_E$, i.e., $$\alpha^n+a_{n-1}\alpha^{n-1}+\dots+a_1\alpha+a_0=0,$$ for $a_0,\dots,a_{n-1}\in\overline F_E$. Let $K=F(a_0,\dots,a_{n-1})$. Clearly $K$ is a finite extension over $F$. The above equation shows that $[K(\alpha):K]\le n$, and thus $$[K(\alpha):F(\alpha)][F(\alpha):F]=[K(\alpha):F]=[K(\alpha):K][K:F]<\infty.$$ Since $[F(\alpha):F]$ is finite, $\alpha$ is algebraic over $F$.