I know there is a formula for the sum of the primitive $ n $th roots of unity which is the Mobius function of $ n $. See: The Möbius function is the sum of the primitive $n$th roots of unity.
I am curious about the existence of a formula for the product of the primitive $ n $th roots of unity.
If $g$ is a primitive $n$-th root of unity, then so is $g^{-1}$.
Unless $n\le2$, we have $g \ne g^{-1}$. When $n=2$, there is only one primitive $n$-th root of unity: $-1$.
Therefore, the product of all primitive $n$-th roots of unity is $1$ when $n\ne 2$ and $-1$ when $n=2$.
This is quite similar to Wilson's theorem for abelian groups:
The product of all elements in a finite abelian group is $1$, except when there is exactly one element of order $2$, in which case the product is this element.
Working out the following definition of the Cyclotomic Polynomial $$ {\displaystyle \Phi _{n}(x)=\prod _{\stackrel {1\leq k\leq n}{\gcd(k,n)=1}}\left(x-e^{2i\pi {\frac {k}{n}}}\right),} $$ you'll get $$ {\displaystyle x^{\varphi(n)}-x^{\varphi(n)-1}\sum _{\stackrel {1\leq k\leq n}{\gcd(k,\,n)=1}}e^{2\pi i{\frac {k}{n}}}} + \dots + 1 $$
For the final $1$, you'll get: $$ \prod _{\stackrel {1\leq k\leq n}{\gcd(k,n)=1}}\left(-e^{2i\pi {\frac {k}{n}}}\right)=\left(e^{\frac{2i\pi}{n} \sum_{\stackrel {1\leq k\leq n}{\gcd(k,n)=1}}k}\right )\prod _{\stackrel {1\leq k\leq n}{\gcd(k,n)=1}}(-1)=\left(e^{\frac{2i\pi}{n}\frac{n}{2}\varphi(n) }\right )(-1)^{\varphi(n)}=+1, $$
where I used this...
That's much easier than the sum! The "primitive nth roots of unity" are $e^{\frac{ik\pi}{n}}$ for k from 1 to n. Their product is $e^{(1+ 2+ ...+ n)(ik\pi)}= e^{\frac{n(n+1)ik\pi}{2}}$.
