The simplest case that can arise (given that p is an odd prime and n has to be an odd integer) is when p=5 and n=3. We then have to form the sum . In this expression, there are four possible triples (ijk), namely (123), (124), (134), (234). In each case i, the smallest integer in the triple, is 1 or 2. So the value of will be +1 every time, and the sum becomes (evaluated mod 5). The inverses of 1,2,3,4 in the field are respectively 1,3,2,4, so the above sum is equal to .
So the result is correct in this case. As soon as the numbers p and n become larger, the number of terms in the sum gets too big for numerical verifications to be any fun.