$\textbf{Q:}$ Is there any obvious reason that $\frac{\prod_{j\leq d}(q^{n+1}-q^j)}{\prod_{k\leq d}(q^{d+1}-q^k)}$ is an integer where $q$ is a power of some prime for all $d\leq n,n\geq 1$?
I used cyclotomic field polynomial and argued that denominator's cyclotomic polynomials must be in the numerator's cyclotomic field polynomials by removing redundant extra factor of $q^j$ and $q^k$. Then the rest is algebraic number and by algebraic closureness of $Z$. I conclude the quotient must be in $Z$.(I need to extend the galois automorphism of cyclotomic fields by considering some large $\zeta_l$ cyclotomic field extension say $l=\prod_{j\leq d}(n+1-j)$.)
This is basically counting distinct $d$ dimension linear subspace of $P_n(F_q)$.
I think this answers the question (but without needing $q$ to be a power of a prime, so I’m not sure): $\displaystyle\frac{\prod_{j\leq d}(q^{n+1}-q^j)}{\prod_{k\leq d}(q^{d+1}-q^k)}=\frac{\prod_{j\leq d}(q^{n+1-j}-1)}{\prod_{k\leq d}(q^{d+1-k}-1)}$, which is a $q$-binomial coefficient, therefore a polynomial.