The complex $m$th roots of unity are roots of $x^m-1$, and form a regular $m$-gon on the unit circle in the complex plane with one vertex at $1$ (for $0 < m \in \mathbb{N}$). Primitive $m$th roots of unity are roots of the irreducible polynomial factors of $x^m-1$, called cyclotomic polynomials $\phi_d(x)$ and satisfying
$$
x^m-1 = \prod_{0$\phi(d)$ and, as stated, its roots are precisely the primitive $m$th roots of unity.
(The polynomial factorization formula above can also be seen as providing a partition of the full set of $m$th roots of unity, which forms a multiplicative subgroup of $\mathbb{C}$ of order $m$, into subsets based on their order -- for which minimal $d$ do they satisfy $x^d=1$ -- comprised of $\phi(d)$ generators $\{e^{2\pi ij/d}|(j,d)=1\}$ of the subgroup of order $d$ for each positive divisor $d$ of $n$. It is worthwhile once to draw some examples of these partitions.)
Now since
$$
[m]_q=\frac{q^m-1}{q-1}=\prod_{11$ (all primitive $d$th roots of unity where $d>1$ and $d \mid m$).
So a root of
$$
\binom{n}{k}_q
=\prod_{j=1}^{k}\frac{[n-j+1]_q}{[j]_q}
=\prod_{1
In particular for $m=d \mid n$, we can see that the primitive $m$th roots of unity are roots of $\binom{n}{k}_q$ iff
$$
\phi_m(q) \mid \binom{n}{k}_q
\iff
e_m = 1
\iff
m \nmid{k}
\iff
m \nmid n-k
\;.
$$