Sum of the polynomial roots raised to a power. How to prove? Problem:
If we have a polynomial $f$ with a derivative $f\,'$ and quotient $q$ function defined as:
$$q(x)=\sum_{i=1}^{\infty}a_ix^{-i}=\frac{f\,'(x)}{f(x)},$$
and the roots of $f$ are $x_1,x_2,\ldots,x_k$, how to prove
$$a_i=\sum_{j=1}^{k}x_j^i$$
Details:
If $f(x)=x^2-5x+6$, $f\,'(x)=2x+5$,
$q(x)=2 x^{-1}+5 x^{-2}+13 x^{-3}+35 x^{-4}+97 x^{-5}+\ldots$
 A: Assume without loss of generality that $f(x)$ is a monic polynomial with $n$ roots $x_1, x_2, \ldots, x_n$ so that we can write
$$f(x) = \prod_{k=1}^n(x-x_k)$$
The product rule for derivatives then gives us
$$f^{\prime}=\frac{\mathrm d}{\mathrm dx}f(x)=\frac{\mathrm d}{\mathrm dx}\prod_{k=1}^n(x-x_k)= \sum_{k=1}^n\ \,\prod_{i=1,i\neq k}^n(x-x_i)$$ 
where the $k$-th term of the sum on the right is the product of all
the $(x-x_i)$ except $(x-x_k)$.  Therefore, 
$$\frac{f^{\prime}}{f} = \frac{\sum_{k=1}^n\prod_{i=1,i\neq k}^n(x-x_i)}{\prod_{k=1}^n(x-x_k)} = \sum_{k=1}^n\frac{1}{x-x_k}.$$
Now, basic "long division" of $1$ by $x-x_k$ produces a "quotient"
$$x^{-1} + x_k\cdot x^{-2} + x_k^2\cdot x^{-3} + \cdots $$
so that 
$$\sum_{k=1}^n\frac{1}{x-x_k} 
= \sum_{i=1}^{\infty}\left(\sum_{k=1}^n x_k^{i-1}\right)\cdot x^{-i}$$
which is essentially the answer wanted by Dan Garou except that,
as noted by Thomas Andrews, it is "off-by-one."  The "long division" can
be formalized by expanding $(1-x_k\cdot x^{-1})^{-1}$ in a  Taylor series 
in $x^{-1}$or the binomial theorem etc. but I will leave the details to Dan 
Garou to fill in.  
Note: If anyone feels strongly enough about the
 cavalier treatment of power series in this last part to 
want to fill in the details, please feel free to 
edit this answer. 
