(Asked in MSE 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 $$ s_{\lambda/\mu}(1_N)=\det\left({N+\lambda_i-i-\mu_j+j-1\choose \lambda_i-i-\mu_j+j}\right).\qquad (1)$$

However, I was reading a paper by Chen and Stanley (A Formula for the Specialization of Skew Schur Functions) and they state that $$s_{\lambda/\mu}(1,q,q^2,...)=\frac{1}{\prod_{u\in\lambda/\mu}[N+c(u)]_q}\det\left(\left\lbrack \begin{matrix} N+\lambda_i-i\\\lambda_i-i-\mu_j+j\end{matrix}\right\rbrack_q\right),\qquad (2)$$ 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 $\left\lbrack \begin{matrix} a\\b\end{matrix}\right\rbrack_q=\frac{[a]_q!}{[b]_q![a-b]_q!}.$

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

since the left hand side of (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?

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

How to obtain equation (1) from equation (2)?