Can you use 2^n - 2 / n to check if a number is prime with 100% accuracy? According to the AKS primality test:
$$(x-1)^p - (x^p-1)$$
If all coefficients (which can be found in Pascal's triangle) are divisible by p then p is prime.
If we sum these coefficients we get:
$2$ for $p = 2$;
$6$ for $p = 3$;
$14$ for $p = 4$;
$30$ for $p = 5$
$\ldots$
If all the coefficients are divisible by p, then the sum of all those coefficients must also be divisible by p
$sum = 2^p - 2$
So if $(2^p - 2) / p$ is a natural number, can we conclude that $p$ is definitely prime?
Please correct me if I made any obvious mistake

If a sum is divisble by $p,$ it does not mean the summands are.
The smallest counterexample to your claim is $p=341.$ We have $341=11\cdot 31,$ but $2^{341}=2\cdot(2^{10})^{34} = 2\cdot(1024)^{34} = 2\cdot(3\cdot 341+1)^{34} \equiv 2\cdot 1^{34} = 2 \pmod{341}.$ 
