(Asked in [MSE](https://math.stackexchange.com/questions/3760125) but got no response.)

The generating function $\frac{1}{(1-t)^N}=\sum_k {N+k-1\choose k}t^k=\sum_k h_k(1)t^k$ and the Jacobi–Trudi formula $s_{\lambda/\mu}=\det(h_{\lambda_i-i-\mu_j+j})$ tell me that the value of the skew Schur function at the identity is
$$ s_{\lambda/\mu}(1_N)=\det\left({N+\lambda_i-i-\mu_j+j-1\choose \lambda_i-i-\mu_j+j}\right).\tag{1}\label{1}$$

However, I was reading a paper by Chen and Stanley ([A Formula for the Specialization of Skew Schur Functions](https://doi.org/10.1007/s00026-016-0312-2)) and they state that 
$$s_{\lambda/\mu}(1,q,q^2,\dotsc)=\frac{1}{\prod_{u\in\lambda/\mu}[N+c(u)]_q}\det\left(\genfrac[]{0pt}{}{N+\lambda_i-i}{\lambda_i-i-\mu_j+j}_q\right),\tag{2}\label{2}$$
where $c(u)$ is the content of the box $u$ in the Young diagram of shape $\lambda/\mu$ and the $q$-quantities are $[x]_q=1-q^x$, $[a]_q!=[a]_q[a-1]_q\cdots$ and $\genfrac[]{0pt}{}a b_q=\frac{[a]_q!}{[b]_q![a-b]_q!}$.

I am not an expert in this $q$-business, and I am confused by this equation. I have a few closely related questions. 

1. Since the left hand side of \eqref{2} is a polynomial in $q$, it should have a limit when $q\to 1$ and this should be the skew Schur at the identity. Is this correct? But what is the number of arguments? 

2. The determinant at the right hand side on \eqref{2} has a limit when $q\to 1$, but the prefactor does not. How to take the limit $q\to 1$ of this equation? 

3. How to obtain equation \eqref{1} from equation \eqref{2}?