I have a rather basic question regarding the $q$-binomial $\begin{bmatrix}N \\ r \end{bmatrix}=\frac{(1-q^N)(1-q^{N-1} ) \dots (1-q^{N-r+1})}{ (1-q)(1-q^2)\dots(1-q^r) }$ as $N$ goes to infinity. On pages 26 and 27 of Macdonald's book on symmetric functions, it says that the $q$-binomial in this limit is given by
\begin{equation} \lim_{N\to \infty }\begin{bmatrix}N \\ r \end{bmatrix} = \frac{1}{ (1-q)(1-q^2)\dots(1-q^r) }~. \end{equation}
My question is for what values of $q$ the above limit for the $q$-binomial holds. While it obviously holds when the absolute value $| q | <1$, it is not specified in Macdonald that this is the case (as far as I can tell), $q$ is simply stated to be indeterminate. Any help is much appreciated.
We can think of this as convergence for formal power series, so the topology here is discrete. This means that the coefficient at each power of $q$ should be eventually constant for the limit to exist. Indeed, $$ \begin{split} \begin{bmatrix}N\\r\end{bmatrix}&=\frac{(1-q^N)(1-q^{N-1})\dots(1-q^{N-r+1})}{ (1-q)(1-q^2)\dots(1-q^r)}\\ &=\frac{1}{(1-q)(1-q^2)\dots(1-q^r)}\cdot(1-q^{N-r+1}+\text{higher powers of $q$}), \end{split} $$ so the power series for $\begin{bmatrix}N\\r\end{bmatrix}$ and for $\frac{1}{(1-q)(1-q^2)\dots(1-q^r)}$ coincide up to power $q^{N-r}$. Letting $N\to\infty$, we get the desired limit.