Let $n,N\in\mathbb{N}$ and $m_1,x_1,\ldots,m_n,x_n\in\mathbb{N}$. I'm trying to rewrite following expression
$$\prod_{j=1}^n\prod_{k=0}^{m_{j}-1} (x_j-kN^{-1})$$
I am only interested in the summands which have a factor $N^0=1$ or $N^{-1}$. So far I was able to calculate for any $j\in\{1,\ldots, n\}$:
$$\prod_{k=0}^{m_{j}-1} (x_j-kN^{-1})=x_j^{m_j}-N^{-1}\binom{m_j}{2}x_j^{m_j-1}+N^{-2}\cdot\ldots$$
Plugging this into the first product, I find
$$ \begin{align} \prod_{j=1}^n\prod_{k=1}^{m_{j}-1} (x_j-kN^{-1})&=\prod_{j=1}^n\big(x_j^{m_j}-N^{-1}\binom{m_j}{2}x_j^{m_j-1}+N^{-2}\cdot\ldots\big)\\ &=\big(\prod_{j=1}^nx_j^{m_j}\big)-N^{-1}\big(\prod_{j=1}^nx_j^{m_j}\big)\sum_{j=1}^n\binom{m_j}{2}x_j^{-1}+N^{-2}\cdot\ldots \end{align}$$
However, is it possible to rewrite the second expression into a nicer form?
$$\big(\prod_{j=1}^nx_j^{m_j}\big)\sum_{j=1}^n\binom{m_j}{2}x_j^{-1} =\ ?$$
Probably rewriting it with a factor $\big(\prod_{j=1}^nx_j^{m_j-1}\big)$, giving an equation like
$$\big(\prod_{j=1}^nx_j^{m_j}\big)\sum_{j=1}^n\binom{m_j}{2}x_j^{-1} = y\cdot \big(\prod_{j=1}^nx_j^{m_j-1}\big)$$
Unfortunately, I do not get any nice result for $y$ neither.
Any help is appreciated! Thank you in advance!