Is there a formula for the coefficient of $x^k$ in the expansion of $\prod_{i=1}^n(ix+1)$

Question

Is there a known formula or can such a formula exist in terms of $k$ and $n$ to find the coefficient of $x^k$ in the expansion of: $$\prod_{i=1}^n(ix+1)$$ For example, the coefficient of $x^{n}$ will be $n!$.

Answer 1

Thanks to the comments, I figured out how to solve this problem. I thought I should write an answer for someone who came later to this post looking for help in this. So, for example, for $n=3$, we have: $$\prod_{i=1}^3(ix+1)$$ which gives $$3!\prod_{i=1}^3(x-(-\frac{1}{i}))$$ So, the roots of the polynomial I'm trying to get are : $-1,\frac{-1}{2}, \frac{-1}{3}.$ Clearly, the coefficient of $x^3$ is 3!=6. By Vieta's formula, the coefficient of $x^2$ will be 3! multiplied by sum of roots. So, coefficient of $x^2$=

$6\cdot (-1+\frac{-1}{2}+\frac{-1}{3})=-11$

Similarly, the coefficient of $x$ will be 3! multiplied by sum of roots taken two at a time which gives:

$6\cdot (-1*\frac{-1}{2}+\frac{-1}{2}*\frac{-1}{3}+\frac{-1}{3}*-1)=6$ And the constant term will be 3! multiplied by product of roots, which gives:

$3!*-1*\frac{-1}{2}*\frac{-1}{3}=-1$