"Oddity" of $q$-Catalan polynomials: Part II

This question extends my earlier MO post for which I'm grateful for answers and useful comments.

The Catalan numbers $$C_n=\frac1{n+1}\binom{2n}n$$ satisfy: $$\text{C_{1,n} is odd iff n=2^j-1 for some j}$$.

The $$2$$-adic valuation of $$x\in\mathbb{N}$$ is the highest power $$2$$ dividing $$x$$, denoted by $$\nu(x)$$. Let $$s(x)$$ stand for the sum of the binary digits of $$x$$. Then, we have the fact that $$\nu(C_{1,n})=s(n+1)-1. \tag1$$ Let $$n+1=n_r2^r+n_{r-1}2^{r-1}+\cdots+n_12+n_0$$ be the binary expansion of $$n+1\in\mathbb{N}$$, for some $$n_j\in\{0,1\}$$. Further, denote by $$(n+1)^*=\{n_{j_1},n_{j_2},\dots,n_{j_t}\}$$ the non-zero digits ordered as $$j_1>j_2>\cdots>j_t$$. Note: $$\#(n+1)^*=s(n)$$.

One version of the $$q$$-Catalan polynomials $$C_n(q)$$ is given in the manner $$C_n(q)=\frac1{[n+1]_q}\binom{2n}n_q;$$ where $$_q:=1, [n]_q=\frac{1-q^n}{1-q}$$ and $$\binom{n}k_q=\frac{[n]_q!}{[k]_q![n-k]_q!}$$. Here $$[n]_q!=_q_q\cdots[n]_q$$.

Working in the spirit of (1), I was curious to find a possible $$q$$-analogue.

QUESTION 1. Is this true? If so, how does the proof go? $$\prod_{k=1}^{t-1} (1+q^{2^{j_k}}) \qquad \text{divides} \qquad C_n(q), \tag2$$ and no other such factors divide it!

REMARK. In view of the fact that the term $$1+q^{2^{j_t}}$$ is absent from the LHS of (2) ensures that (2) indeed emulates (1), naturally.

For a positive integer $$m$$, an $$m$$-th primitive root $$\alpha_m$$ of $$-1$$ [which is the root of the polynomial $$1+q^m$$, and when $$m$$ is a power of two the converse also holds: any root of $$q^m+1$$ is an $$m$$-th primitive root of $$-1$$] is a root of $$[k]_q$$ exactly when $$2m$$ divides $$k$$. So, $$\alpha_m$$ is a root of $$[N!]_q$$ of multiplicity $$[\frac{N}{2m}]$$. Therefore, $$\alpha_m$$ is a root of $$C_n(q)$$ of multiplicity $$\left[\frac{2n}{2m}\right]-\left[\frac{n}{2m}\right]-\left[\frac{n+1}{2m}\right].$$ When is this positive? For $$m=1$$ never, so assume that $$m>1$$. Denote $$n+1=2mk+r$$, $$0\leqslant r<2m$$. Then we have $$\left[\frac{2n}{2m}\right]-\left[\frac{n}{2m}\right]-\left[\frac{n+1}{2m}\right]= 2k-\chi_{r=0}+\chi_{r>m}-\left(2k-\chi_{r=0}\right)-2k=\chi_{r>m}.$$ Now if $$m=2^a$$, and $$n+1=2^{j_1}+2^{j_2}+\dots+2^{j_t}$$, $$j_1>j_2>\dots>j_t$$, the remainder of $$n+1$$ modulo $$2^{a+1}$$ is greater than $$2^a$$ exactly when $$a\in \{j_1,j_2,\dots,j_{t-1}\}$$. This implies your claim.
Note also that $$C_n(q)$$ is a product of cyclotomic polynomials in certain powers (since so is each $$[k]_q$$). All cyclotomic polynomials other than $$\Phi_{2^{a+1}}=1+q^{2^a}$$, $$a=0,1,\dots$$, take value 1 at point 1. Therefore substituting $$q=1$$ to the $$q$$-version we get the usual $$q=1$$ oddity statement.