TITLE: Some curious Hankel determinants
QUESTION [5 upvotes]: Let $f(n,q)=\prod_{j=1}^na(q^j)$ for a polynomial $a(q)$ and let $d(n)=\det(f(i+j,q))_{i,j=0}^n$ be its Hankel determinant.
Computer experiments suggest that
$$\lim_{q\to1}\frac{d(n)}{(q-1)^\binom{n+1}{2}}=(a(1)a'(1))^\binom{n+1}{2}\prod_{j=0}^nj!.$$
Has anyone an idea how to prove this?
Remark: For $a(q)=a+qb$ it is easy to verify that
$$d(n)=((q-1)b)^\binom{n+1}{2}q^\frac{n(n+1)(2n+1)}{6}{\prod_{j=1}^n[j]_{q}!(a+q^jb)^{n+1-j}},$$
if $[n]_{q}=\frac{1-q^n}{1-q}$ and $[n]_q!=[1]_q[2]_q\dots[n]_q.$
Therefore the conjecture is true for linear polynomials and also for $a(q)=q^m.$

REPLY [5 votes]: The identity is clearly invariant to scaling $a$ by a constant. (One has to check the constant appears in the determinant to power $n(n+1)$). So wlog we may assume $a(1)=1$. 
We can expand $\log a$ as a formal power series in $\log q$, $$ a(q)= e^{ \sum_{i} a_i (\log q)^i}$$ $$a (q^j) = e ^{ \sum_i a_i  j^i (\log q)^j}$$ $$\prod_{j=1}^n  a(q^j)  = e^{ \sum_i a_i (\sum_{j=0}^n j^i) (\log q)^j }$$
the key thing being that $(\sum_{j=0}^n j^i)$ is a polynomial in $n$ of degree $\leq n+1$.  Hence if we expand this product $f(n,q)$ out as a power series in $\log q$, the coefficient of the $i$th power of $\log q$ will be a polynomial in $n$ of degree $\leq 2i$, whose leading term is a power of $a_1$.  
Hence the coefficient of $\log q^i$ in the power series of $f(x+y,q)$ is a sum of monomials in $x$ times monomials in $y$ of total degree $\leq i$. 
When we take a determinant of a sum of rank one matrices, in this case a sum indexed by powers of $\log q$ and then pairs of monomials, it's the same as the sum of the determinants of all sums of $n+1$ matrices from the set. These determinants are nonzero only if the degreees of all the monomials in $x$ and monomials in $y$ appearing are distinct, which means the total degree in each variable must be at least ${n+1 \choose 2}$, for a total degree of at least $2 {n+1  \choose 2}$, and thus total degree in $\log q$ of at least ${n+1 \choose 2}$. This is only sharp if the monomials we study are the highest degree terms, which we saw come only from $a_1$.
So the answer depends only on $a_1$, and the calculations you include in your answer finish the job.