A question about q-binomials at roots of unity

I have a question about a lemma $9.3.6$ in the book A Guide to Quantum Groups written by Vyjayanthi Chari and Andrew Pressley. This question comes from page 301, "The restricted specialization".

Here is the lemma:

Let $\epsilon$ be a primitive $\ell$th root of unity, where $\ell$ is odd and greater than one. Let $r,s\in \mathbb{N}$ and write $r=r_0+\ell r_1,s=s_0+\ell s_1$, where $0\le r_0,s_0<\ell$, $r_1,s_1\ge 0$, $r\ge s$. Then $$\left[ \begin{array}{c} r\\ s\\\end{array} \right] _{\varepsilon}=\left[ \begin{array}{c} r_0\\ s_0\\\end{array} \right] _{\varepsilon}\left( \begin{array}{c} r_1\\ s_1\\\end{array} \right).$$where $\left( \begin{array}{c} r_1\\ s_1\\\end{array} \right)$ is an ordinary binomial coefficient

Here are some notations:\begin{align}[n]_q& =\frac{q^n-q^{-n}}{q-q^{-1}}, \\[n]_q!& =[n]_q.[n-1]_q\cdots [2]_q.[1]_q, \\\left[\begin{array}{c}m\\n\end{array}\right]_q& =\frac{[m]_q!}{[n]_q![m-n]_q!}. \end{align}

Here is my question:

Since $\epsilon$ is a primitive $\ell$th root of unity, then I think $[l]_\varepsilon$ equals to $0$.

So the denominator will have $0$. Does this lemma make sense?

I am confused about it.

Any help and references are greatly appreciated.

Thanks!