Is the following a valid proof?
Let $a = g^k$ mod $n$ where $g$ is a primitive root of $n$
$a^n = (g^k)^n \text{mod} (n) = (g^n)^k \text{mod }(n) = g^k \text{mod}(n)$ [as $g^n = g$ mod $n$] $= a$ mod $n$
Same for $a^{n-1} = 1$ mod $n$ where $\text{gcd}(a,n)= 1$
let $a = g^k$
$a^{n-1} = (g^k)^{n-1} \text{mod } n = (g^{n-1})^k \text{mod } n = 1^k \text{mod } n = 1 \text{mod } n$
Well, the problem is not so much the correctness of your proof - it's correct, but using a primitive root to prove Fermat's little theorem is weird. I seriously doubt that you can prove the existence of primitive roots without assuming Fermat's theorem, or at least without using some statement at some point directly implying it.
It's like proving that if $a\wedge n = 1$ then $a$ is invertible modulo $n$ saying "by little Fermat's theorem, $a^{n-1} = 1 \mod n$, so $aa^{n-2} = a^{n-2}a = 1 \mod n$". The sentence is true, the assertions are true, but this is not really a valid proof because you can't really prove Fermat's theorem without using inverses of elements at some point.