I was trying to understand the derivation of
$$ s_{\lambda}=a_{\lambda+\delta} / a_{\delta}=q^{n(\lambda)} \prod_{x \in \lambda} \frac{1-q^{n+c(x)}}{1-q^{h(x)}} $$
as given in (https://math.berkeley.edu/~corteel/MATH249/macdonald.pdf) Page 44, where $b(\lambda)=n(\lambda)= \sum (i-1)\lambda_i$. Here, $s_\lambda$ is the Schur function.
The hook length $h(x)= \lambda_i+\lambda^{\prime}_j-i-j+1$, where $\lambda_i$ is the length of row $i$ and $\lambda^{\prime}_j$ is the length of column $j$. c(x) is the content which is equal $j-i$
I properly understood till the point they reached
$$ \begin{aligned} a_{\lambda+\delta} =\prod_{i<j} q^{\lambda_j+n-j} \left(1-q^{\lambda_{i}-\lambda_{j}-i+j}\right). \end{aligned} $$
After this, I am unable to see how
$$ \frac{q^{n(\lambda)+n(n-1)(n-2) / 6} \prod_{i>1} \varphi_{\lambda_{i}+n-i}(q)}{\prod_{x \in \lambda}\left(1-q^{h(x)}\right)} $$
is obtained, where $\varphi_{r}(q)=(1-q) \ldots\left(1-q^{r}\right)$. Even after this, how do we arrive at the formula.
Can someone please help me out.