Please tell me if my logic is correct.
I know that the Galois group divides the order of the extension and will use this fact as given.
If $K$ is the splitting field for irreducible $f(x)$ over $F$, there are $n$ roots for $f$ in $K$. Then our Galois group could be made out of isomorphism that keep $F$ fixed by sensing roots to roots. So for instance if $a_1,a_2...a_3$ are roots of $f$, our first automorphism could send $a_1$ to $a_2$ and this gives us $n!$ as our group order at the very least. And $K:F$ is a extension of less than $n!$ so order Galois group is equal to the order of extension and we are done.
My book wrote 2 pages for this proof and this makes me feel like my proof is missing something/wrong Is this wrong? Please help. (I am new to this area so please forgive me)
The elements of the Galois group is not merely "permutations of roots". They are automorphisms of the field.
To give an example: let $F$ be the rational field $\Bbb Q$, and let $f$ be the polynomial $x^4 - 10 x^2 + 1$.
Then $f$ has $4$ roots: $\sqrt 2 + \sqrt 3$, $\sqrt 2 - \sqrt 3$, $- \sqrt 2 + \sqrt 3$, $- \sqrt 2 - \sqrt 3$.
Not all the $24$ permutations of these $4$ roots are automorphisms of the splitting field. To see this, suppose that a field automorphism sends $\sqrt 2 + \sqrt 3$ to $\sqrt 2 - \sqrt 3$, then it must send $-\sqrt 2 - \sqrt 3 $ to $-\sqrt 2 + \sqrt 3$.
In other words, you don't have "total freedom" to permute these roots.
Does this answer your question?