How to interpret this q-binomial limit

Question

In the Cousera course on enumerative combinatorics the lecturer claims:

$\lim_{m\to\infty} \begin{bmatrix} m+n\\ n \end{bmatrix}_q = \lim_{m\to\infty}\frac{(1-q^{m+1})\cdot(1-q^{m+2})\cdots(1-q^{m+n})}{(1-q)\cdot(1-q^2)\cdots(1-q^n)} = \frac{1}{(1-q)\cdot(1-q^2)\cdots(1-q^n)}$.

Now clearly in a conventional analysis sense this is fine for $|q|<1$, but we are dealing with formal power series here, and the result is meant to hold for the formal variable $q$. Indeed the left hand side can be thought of as the generating function for Young diagrams fitting into a rectangle with width $n$ and infinite height which agrees with the right hand side, which is the generating function for partitions into a maximum of $n$ summands.

My question is, how can we understand this limit?

Answer 1

$\left(1-q^{m+1}\right)\cdot\left(1-q^{m+2}\right)\cdots\left(1-q^{m+n}\right) = 1 -q^mP(q),\text{ with } P\in\mathbb{Z}[q]$ so the limit is $1$ for the $q$-adic topology of $\mathbb{Z}[[q]]$

Answer 2

For power series, the phrase $\lim_{n\to\infty} f_n(q)=f(q)$ means that, for all $k\in \mathbb N$, there exists an $N\in \mathbb N$ so that $$ n\ge N\implies [q^k]f_n(q)=[q^k]f(q) $$ In words, for each $k$, the coefficient of $q^k$ in $f_n(q)$ must eventually match the coefficient of $q^k$ in $f(q)$. Since $$ {n+m \brack n}_q=\frac{1-q^{m+1}+(\text{higher powers of }q)}{(1-q)\cdots(1-q^n)} $$ it follows that the first $m$ coefficients of ${n+m \brack n}_q$ will agree with that of $\frac1{(1-q)\cdots(1-q^n)}$.