Let $S \subset \mathbb{N}$ and let $\Phi_S = \{ \Phi_n(q) \mid n \in S \}$ where $\Phi_n(q)$ denotes de $n$-th cyclotomic polynomial. Denote by $\Phi_S^*$ de multiplicative subset associated to $\Phi_S$ ordered by divisibility. Then we can consider the completion $$\mathbb{Z}[q]^S = \varprojlim_{f \in \Phi_S^*} \mathbb{Z}[q]/(f)$$
Why when considering $S = \mathbb{N}$ we get $$\mathbb{Z}[q]^{\mathbb{N}} = \varprojlim_{n\in \mathbb{N}} \mathbb{Z}[q]/((q)_n)$$ where $$(q)_n = (1-q)(1-q^2)\cdots (1-q^n)$$
The question really is why $f(q) \in \Phi_{\mathbb{N}}^*$ is of the form $(q)_n$ for some $n$. I believe I could answer this question myself with some better knowledge of the properties of cyclotomic polynomials, so if someone could point me out and article or something where this things are explained, I'd be grateful too. Thank you!
Edit: I believe I didn't put much thought into this question. Obviously it is not true that every $f \in \Phi_{\mathbb{N}}^*$ can be written as $(q)_n$ for some $n \geq 0$. What is true however, is that for each polynomial $f \in \Phi_{\mathbb{N}}^*$, there is some $n$ such that $f \mid (q)_n$. That means that the set $\{(-1)^n(q)_n \mid n \geq 0\}$ is cofinal in $\Phi_{\mathbb{N}}^*$, and hence $$\mathbb{Z}[q]^S = \varprojlim_{f \in \Phi_S^*} \mathbb{Z}[q]/(f) \cong \varprojlim_{n\in \mathbb{N}} \mathbb{Z}[q]/((q)_n)$$ as desired.