TITLE: Prove that the expression is a whole number
QUESTION [6 upvotes]: Let $\;q\in\mathbb{N}\;,\;q\geqslant2\;$ and $\;n\in\mathbb{N}\;,\;n\geqslant1.$
Prove that for all $\;m\in\mathbb{N}$ :
$\alpha_m=\dfrac{(q^m-1)(q^m-q)\cdot\ldots\cdot(q^m-q^{n-1})}{(q^n-1)(q^n-q)\cdot\ldots\cdot(q^n-q^{n-1})}\in \mathbb{N}\;.$
There is a hint given: start by dealing with the case $\;q=p^i\;$ with $\;p\;$ prime and $\;i\geqslant1.$
I have tried a few routes ( including induction starting with $\;m\geqslant n\;$ because $\;m<n\;$ yields $\;0\;$ straight away ), but I cannot quite bridge every gaps. This question is from “Agregation externe 2021 : question III-12-a”.
By induction, I have tried to prove that:
$\alpha_m \in \mathbb{N}$ and is a multiple of $(q^{m-n}+q^{m-n-1}+...+1)$
$\alpha_n=1$, and I end up with $\alpha_{m+1}=\frac{q^m+q^{m-1}+...+1}{q^{m-n}+q^{m-n-1}+...+1}\alpha_m$, so $\alpha_{m+1} \in \mathbb{N}$ but I am missing the second part of the induction hypothesis.
Also, I am not using the hint, which makes me think this is not the right way to proceed...

REPLY [4 votes]: Hint
Denote
$$\binom{m}{n}_q= \dfrac{(q^m-1)(q^m-q)\cdot\ldots\cdot(q^m-q^{n-1})}{(q^n-1)(q^n-q)\cdot\ldots\cdot(q^n-q^{n-1})}.$$
$\binom{m}{n}_q$ is called the Gaussian binomial coefficient.
One can prove the relations
$$\binom{m}{n}_q= \frac{q^m-1}{q^n-1}\binom{m-1}{n-1}_q$$ and
$$\binom{m}{n}_q=q^n \binom{m-1}{n}_q+ \binom{m-1}{n-1}_q$$
enabling to prove that those numbers are natural numbers by induction.