Compute$\sum\limits_s\frac{s^{n+1}}{s-1}$ where the sum enumerates the $k$th roots $s$ of $2$ 
I am trying to prove that
  $$\sum_{r=0}^{k-1} \frac{(2^{1/k}w^r)^{n+1}}{2^{1/k}w^r-1}=2^{\lfloor n/k\rfloor +1} k$$
  for integer $n\geq 0$, where $w$ is the primitive $k^{th}$ root of unity (meaning that $2^{1/k}w^r$ for $r=0,\dots,k-1$ are the $k^{th}$ roots of two) and $\lfloor x \rfloor$ (the floor function) gives the integer part of $x$.

Having tried this out numerically I'm confident this is the answer, but finding it hard to prove.
I have tried writing $w=\cos(2 \pi /k)+i\sin(2\pi /k)$, but things get messy quickly.
I've also tried re-writing as 
$$\sum_{r=0}^{k-1}(2^{1/k}w^r)^{n} + \frac{(2^{1/k}w^r)^{n}}{2^{1/k}w^r-1}$$
but although the first term is simple enough to analyze I have the same sorts of problems with the second as I had originally due to the roots on the bottom.
Any ideas would be gratefully received.
 A: First cosider the summand: 
$$\frac{(2^{1/k}w^r)^{n+1}}{2^{1/k}w^r-1}$$
which can be re-written as 
$$\frac{(2^{1/k}w^r)^{n}}{1-2^{-1/k}w^{-r}}$$
which we can further recognise as the solution to a geometric sum:
$$\displaystyle\sum_{j=0}^{\infty}(2^{1/k}w^r)^{n}(2^{-1/k}w^{-r})^{j}\\
=\displaystyle\sum_{j=0}^{\infty}(2^{1/k}w^r)^{n-j},\\
=\displaystyle\sum_{j=-\infty}^{0}(2^{1/k}w^r)^{n+j},\\
=\displaystyle\sum_{j=-\infty}^{n}(2^{1/k}w^r)^{j}.$$
Hence 
$$\sum_{r=0}^{k-1}\frac{(2^{1/k}w^r)^{n+1}}{2^{1/k}w^r-1}=\sum_{r=0}^{k-1}\displaystyle\sum_{j=-\infty}^{n}(2^{1/k}w^r)^{j},\\
=\displaystyle\sum_{j=-\infty}^{n}2^{j/k}\sum_{r=0}^{k-1}w^{rj},\\
=\displaystyle\sum_{j=-\infty}^{n}2^{j/k}k\mathbb{I}_{k|j},$$
where $\mathbb{I}_{k|j}$ is unity if  $\quad j\mod k=0$ and zero otherwise.
In the penultimate line we have used
$$\sum_{r=0}^{k-1}w^{rj}=k\mathbb{I}_{k|j},$$ a propert of the $k^{th}$ roots of unity.
Letting $j=kl$ for integer $l\leq n/k$ gives
$$\sum_{r=0}^{k-1}\frac{(2^{1/k}w^r)^{n+1}}{2^{1/k}w^r-1}=\sum_{l=0}^{\lfloor n/k \rfloor} 2^l=2^{\lfloor n/k \rfloor+1}k.$$
