Show that if $[E:F]$ is finite then $[E(u):E]\leq [F(u):F]$ is a finite extension

Question

Suppose that $F\subseteq E \subseteq K$ and $[E:F]$ is finite and $u\in K$ is algebraic over $F$.

I am trying to show that $[E(u):E]\leq [F(u):F]$. I have been trying to use the degrees law but nothing so far has worked. How should I start?

Answer 1

$|E(u):E|=\deg f$ where $f$ is the minimum polynomial of $u$ over $E$.

$|F(u):F|=\deg g$ where $g$ is the minimum polynomial of $u$ over $F$.

How are the polynomials $f$ and $g$ related?