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I have the following equations: \begin{aligned} (x+x_1)(x+x_2)&=x^2+(x_1+x_2)x+x_1x_2\\ (x+x_1)(x+x_2)(x+x_3)&= x^3+(x_1+x_2+x_3)x^2\\ &\ \ \ \ \ \ \ \ \ \ +(x_1x_2+x_1x_3+x_2x_3)x+x_1x_2x_3\\ (x+x_1)(x+x_2)(x+x_3)(x+x_4)&=x^4+(x_1+x_2+x_3+x_4)x^3\\ &\ \ \ \ \ \ \ \ \ \ +(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_3+x_3x_4)x^2\\ &\ \ \ \ \ \ \ \ \ \ +(x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4)x\\ &\ \ \ \ \ \ \ \ \ \ +x_1x_2x_3x_4\\ &\ \ \vdots\\ (x+x_1)(x+x_2)\cdots(x+x_{n-1})(x+x_n)&=c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0 \end{aligned} Where $x_1,\dots,x_n$ are known. My question is if there is a general formula to calculate all of the coefficients $c_i$. At the moment I only know that $c_n=1$ and $c_0=\prod\limits^n_{i=1}x_i$. Thanks in advance.

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Yes, in fact. The generating function of the elementary symmetric polynomials in $n$ variables is $$S_n(x)=\prod_{i=1}^n{(1+x_ix)}=\sum_{i=0}^n{e_i(x_1,\ldots,x_n)x^i}$$ where $$e_0(x_1,\ldots,x_n)=1$$ and $$e_i(x_1,\ldots,x_n)=\sum_{1\leq p_1<\cdots<p_i\leq n}{x_{p_1}\cdots x_{p_i}}$$ It's simply the sum over all subsets of the variables with $i$ elements of the product of the variables in the subset.

This is not quite your product. Your product is $$x^nS_n\left(\frac1x\right)$$ Thus the coefficient of $x^{n-i}$ is $e_i$.

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