I observed that if $n$ has a primitive root then
$c_1 \cdot c_2 ... \cdot c_{\phi(n)} \equiv -1 \ mod \ n$
otherwise,
$c_1 \cdot c_2 ... \cdot c_{\phi(n)} \equiv 1 \ mod \ n$
where $c_i$'s are coprime numbers to n smaller than n and $\phi(n)$ is Euler function.
My question is, how to prove that with using primitive roots?
Cases $n=1,2$ are trivial, so let's assume $n > 2$.
Split $\{c_i\}$ in two sets: $\{a_i\}$ of roots of $1$ modulo $n$ and $\{b_i\}$ of non-roots of $1$.
Clearly, $\prod_i b_i \equiv 1 \pmod n$ since inverse modulo $n$ is unique when exists.
And $\{a_i\}$ is breaking into pairs $a_i$ and $-a_i$, so $\prod_i a_i \equiv (-1)^{r(n)/2}$, where $r(n)$ denotes number of roots of $1$ modulo $n$.
So $c_1 \cdot \ldots \cdot c_1 \equiv 1$ iff $r(i) \equiv 0 \pmod 4$.
From identity $a^2-1 = (a-1)(a+1)$ follows that $r(i)=2$ for $n=4$, $p^k$ or $2p^k$, where $p$ is odd prime. For all other $n > 2$ we have $r(i) \equiv 0 \pmod 4$: if $n$ has odd prime factor, then it's by multiplicativity ($r(mn)=r(n)r(m)$ for coprime $m,n$), and for $n=2^k$ with $k > 2$ there is always only roots $\pm 1, 2^{k-1}\pm1$ (it can bee seen from identity for $a^2-1$ again).
Obviously, when $r(n) > 2$ there is no primirive root modulo $n$. And it is well-known that for $n$'s mentioned above with $r(n)=2$ primitive roots exists.
