Show that the product of the $\phi(p-1)$ primitive roots of $p$ is congruent modulo $p$ to $(-1)^{\phi(p-1)}$ 
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.
 A: 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 $p=2$.
A: We know  $ ord_m(a^k) =\frac{d}{(d, k)} $ where d=$ord_ma$ => $ord_pa=ord_p(a^{-1})$
There are $\phi(p-1)$ primitive roots for prime $p$.
As $\phi(n)$ is even for $n>2$ ,i.e. for $p-1>2$.
So,for $p>3$, we can always find  $a^{-1}$ for each primitive root $a$ of $p$.
Now if $a≡a^{-1}(mod\ p)$ =>$a^2≡1(mod\ p)$ =>$ord_pa|2$.
But $\phi(p-1)$>2 for n>6.
So, there will be $\frac{\phi(p-1)}{2}$  pairs each having product $≡1(mod\ p)$ if $p>3$.
The product is ≡$(-1)^{\phi(p-1)}\pmod p$ ,  for $p>3$ .
For $p=3$, the only primitive root is $=2 ≡-1(mod 3)=(-1)^{(\phi(3)-1)}\,$ ,  as $\,\phi(3)=2$
A: We know  $ ord_m(a^k) =\frac{d}{(d, k)} $ where d=$ord_ma$ => $ord_ma=ord_m(a^{-1})$  where m is a natural number.
So, $a,a^{-1}$ must belong to the same order(d).
Now by the previous solution, $a≢a^{-1}(mod\ m)$ if d>2.
So, the product of all number belonging to the same order(d)≡1(mod m) if d>2.
Now, $\phi(m)>2$ if m=5 or >6.
Now, $\phi(m)=2$ if m=3,4,6.
In all the three cases, the primitive root is (m-1)≡-1(mod m).
Here m does not need to be prime and $2<d≤\phi(m)$. 
