(i) Suppose $F \subset E_2 \subset L$ and suppose $\alpha \in L$. Can you show that $[E_2(\alpha):E_2] \leq [F(\alpha):F]$?
[Hint: $[F(\alpha):F]$ is the degree of the minimal polynomial of $\alpha $ over $F$, and $[E_2(\alpha):E_2]$ is the degree of the minimal polynomial of $\alpha$ over $E_2$. How are the degrees of these minimal polynomials related?]
(ii) Let $E_1, E_2$ be finite extensions of $F$, all contained inside $L$. Convince yourself that $E_1 = F(\alpha_1, \dots, \alpha_n)$ for certain $\alpha_1, \dots, \alpha_n \in L$. Convince yourself that $E_1E_2 = E_2(\alpha_1, \dots, \alpha_n)$. Can you use the result of (i) inductively to show that $[E_1 E_2 : E_2] \leq [E_1 : F]$?
[Of course, a similar argument gives $[E_1 E_2 : E_1] \leq [E_2 : F]$.]
(iii) Apply the tower law to the inclusions $F \subset E_1 \subset E_1 E_2$ and $F \subset E_2 \subset E_1 E_2$. Use the fact that $[E_1 : F]$ and $[E_2 : F]$ are relatively prime, together with the result from (ii), to show that $[E_1 E_2 : E_2] = [E_1 : F]$ and $[E_1 E_2 : F] = [E_1 : F][E_2 : F]$.