TITLE: Show that the product of the $\phi(p-1)$ primitive roots of $p$ is congruent modulo $p$ to $(-1)^{\phi(p-1)}$
QUESTION [6 upvotes]: If $p$ is a prime, show that the product of the $\phi(p-1)$ primitive roots of $p$ is congruent modulo  $p$ to $(-1)^{\phi(p-1)}$.

I know that if $a^k$ is a primitive root of $p$ if gcd$(k,p-1)=1$.And sum of all those $k's$ is $\frac{1}{2}p\phi(p-1)$,but then I don't know how use these $2$ facts to show the desired result.
Please help.

REPLY [4 votes]: If $a^k$ is a primitive root modulo $p$, then so is $a^{-k}$. Thus, if $p-1$ is even, then the sum of coprime integers between 1 and $p-1$ with $p-1$ must be =$\phi(p-1)(p-1)/2$, and hence the product of $\phi(p-1)$ primitive roots modulo p must be $a^{\phi(p-1)(p-1/2)}\equiv(-1)^{\phi(p-1)} \pmod p$. If $p-1$ is odd, then the result is trivial, as $p=2$.