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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 $n!$.

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    Do you have a typo? As written, $\prod_{i=1}^n (ix+1)^n=(1+ix)^n.$ – 2017-02-26
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    see Vieta's formulas relating the coefficients of a polynomial to elementary symmetric functions of its roots (in this case, $-1/i$ for $i\in\{1,2,3,\dots,n\}$) – 2017-02-26
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    Write $\prod_{i=1}^n(ix+1)=n!\prod_{i=1}^n (x-(-1/n))$ and use[Vieta's Formulae](https://en.wikipedia.org/wiki/Vieta's_formulas) – 2017-02-26
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    @Dr.MV How is $\prod_{i=1}^n(ix+1)=n!\prod_{i=1}^n (x-(-1/n))$. And $(x-(-1/n))$ doesn't even contain $i$. Then what am I supposed to vary from $1$ to $n$? Please write an answer. Please. – 2017-02-26
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    That should read $x-(-1/i)$ obvioulsy. Have you reviewed the link on Vieta's Formulae? – 2017-02-26
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    @Dr.MV Yes.I followed the link. How am I supposed to use Vieta's Formulae in here? I'm not very good in maths. Would you please write an answer? I need help in this. – 2017-02-26
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    Pick a value of $n$, work out the coefficients, then look `em up in the Online Encyclopedia of Integer Sequences. You will find that you have a version of the Stirling numbers. Look those up, and post an answer to let us know what you have learned. – 2017-02-26
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    Have you tried following up on my suggestion, Dove? – 2017-02-27
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    @GerryMyerson Yeah, I figured out how to do it. I realized that I wasn't making any effort. – 2017-02-28
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    Good. Let me encourage you to write up what you figured out, and to post it as an answer. – 2017-02-28
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    @GerryMyerson I posted it. I got 6, -11, 6, -1 for $n=3$. I looked it up on the online encyclopedia that you told me about. There were many matches. One of them was: "Triangle read by rows of Stirling numbers of first kind, s(n,k), n >= 1, 1<=k<=n." Were you talking about this? – 2017-02-28
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    Yes, that's it. – 2017-02-28

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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:

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