Suppose that n has a primitive root. Show that $\displaystyle \prod _{1 \leq d \leq n } d \equiv -1 \ (mod \ n) $, where gcd(d,n) = 1.
Let $\displaystyle r$ be the primitive root for $\displaystyle n$ then it means $\displaystyle \prod_{d=1}^n d \equiv \prod_{k=1}^n r^k = r\cdot r^2 \cdot ... \cdot r^n = r^{n(n+1)/2}(\bmod n)$.