Let $F$ be a field and let $\alpha_1, \ldots, \alpha_n$ be algebraic over $F$. Prove that $F(\alpha_1,\ldots,\alpha_n)$ is the composite of the fields $F(\alpha_i)$.
Recall that the set $K$ of all algebraic elements over $F$ is a field, and we consider finite extensions to be subfields of $K$. Then the composite of the fields $E_1,\ldots,E_n$ is the inclusion-smallest subfield of $K$ containing all the $E_i$. (The intersection of subfields is a subfield, so this is just the intersection over the class of subfields containing the $E_i$.)
Certainly $F(\alpha_1,\ldots,\alpha_n)$ is a subfield containing each of the $F(\alpha_i)$. Now if $E$ is a field containing the $F(\alpha_i)$, then $E$ contains $F(\alpha_1,\ldots,\alpha_n)$ by definition. ($F(A)$ is the smallest subfield containing $F$ and $A$, for any set $A$.)
So $F(\alpha_1,\ldots,\alpha_n)$ is indeed the composite of the fields $F(\alpha_1),\ldots,F(\alpha_n)$.