# Generated field is the same as composite field

I am considering Exercise 13.2.6 from Dummit & Foote's Abstract Algebra:

Prove directly from the definitions that the field $F(\alpha_1, \alpha_2, \dots, \alpha_n)$ is the composite of the fields $F(\alpha_1,), F(\alpha_2), \dots, F(\alpha_n)$.

We want to prove from definitions that $F(\alpha_1, \alpha_2, \dots, \alpha_n) = F(\alpha_1,), F(\alpha_2), \dots, F(\alpha_n)$

I want to proceed by using minimality of the generated field and composite field. By definition, $F(\alpha_1, \dots, \alpha_n)$ is the smallest field which contains $F$ as well as $\alpha_1, \dots, \alpha_n$ and therefore $F(\alpha_1,\dots,\alpha_n) \subseteq F(\alpha_1)\dots F(\alpha_n)$. Similarly, the composite $F(\alpha_1)\dots F(\alpha_n)$ is the smallest field containing each of $F$ and the $\alpha_i$'s so we have the reverse inclusion $F(\alpha_1)\dots F(\alpha_n) \subseteq F(\alpha_1\dots \alpha_n)$ which completes the proof.

I am unsure about this proof because it seems to be rather obvious and straightforward. Sometimes when I've felt this way, it was because I was assuming some important step that required proof. I want to get the community's thoughts: Is there a different way of approaching the problem? Is this one complete?

$F(\alpha_1,\cdots,\alpha_n)$ contains $F$ and $\alpha_i$ or $i=1,2,3,\dots,n$, so $F(\alpha_i)\subseteq F(\alpha_1,\cdots,\alpha_n)$, which, by the definition of composite, means $F(\alpha_1)\cdots F(\alpha_n)\subseteq F(\alpha_1,\cdots,\alpha_n)$.
On the other hand, $F(\alpha_1)\cdots F(\alpha_n)$ contains $F(\alpha_i)$, by the definition of composite, and hence contains $F$ and $\alpha_i$ for each $i=1,2,\dots,n$, which, by definition, means $F(\alpha_1,\cdots,\alpha_n)\subseteq F(\alpha_1)\cdots F(\alpha_n)$ by the definition of $F(\alpha_1,\cdots,\alpha_n)$.