Error in Proof of Residues? I wanted to prove that the function  
$$F(z) = \frac{z-\sum_{j =2}^{n-1} z^j}{1-\sum_{k=1}^{n}  z^k} $$ 
will only contain simple poles. Is the following proof correct?  

Which implies that $z_o$ has order $1$.
Since I know that this proof isn't fully true, does anyone have any suggestions? Or new methods of proving it?

The question is whether the polynomial $ 1 - \sum_{k=1}^n z^k = \dfrac{z^{n+1} - 2 z + 1}{1-z}$ has a multiple root.  Since $z=1$ is not a multiple root, this is equivalent (for $n>1$) to whether $z^{n+1} - 2 z + 1$ has a multiple root, i.e. whether there is
$z$ with $z^{n+1} - 2 z + 1 = 0$ and $(n+1) z^n - 2 = 0$.  But if $z^n = \dfrac{2}{n+1}$ we have
$$z^{n+1} - 2 z + 1 = z \left(\frac{2}{n+1} - 2\right) + 1 = \dfrac{-2n}{n+1} z + 1$$ so this would say $z =  \dfrac{n+1}{2n}$.
If $p$ is a prime dividing $n$, then $\dfrac{n+1}{2n}$ is a rational number whose denominator (in lowest terms) is divisible by $p$, and so is  any positive integer power of $\dfrac{n+1}{2n}$, but that is not the case for $2/(n+1)$ (note that $n+1$ and $n$ can't both be divisible by the same prime).
