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 $[0]_q:=1, [n]_q=\frac{1-q^n}{1-q}$ and $\binom{n}k_q=\frac{[n]_q!}{[k]_q![n-k]_q!}$. Here $[n]_q!=[1]_q[2]_q\cdots[n]_q$.

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.