Question on a step in a proof regarding complex roots of unity powers Given that I have the expression 
$$N^{k-1}k!\sum_{n=0}^{N-1}w_N^{-nk}e^{\frac{w_N^{n+m}x}{N}}$$
and given that $N=2m$, The next line of the proof states that the above is equal to 
$$N^{k-1}k!\sum_{n=0}^{N-1}w_N^{-nk}w_N^{mk}e^{\frac{w_N^{n}x}{N}}$$
Now i know that we have $-w_N^n=w_N^{n+m}$ given that $N$ is even.  this is just negative complex numbers which rotate complex numbers 180 degrees.  But why, given what I know here is that move okay?
EDIT:  I think that the reason it is true is based upon the fact that $N=2m$.  Thus, when summing, your new starting point is now the root of unity exponent that is opposite the original root of unity exponent.  You are still summing over the same exponentials, but factoring out a $w_N^m$.  Does that seem reasonable?
 A: Based upon the fact that $w_N=e^{\frac{2i\pi}{N}}$, look at the summation and forget the coefficient.  Also take note that since $w_N^N=1, w_N^m=-1$ since $N=2m$.  Now, you just factor out $w_N^{mk}$...
\begin{align}
\sum_{n=0}^{N-1}w_N^{-nk}e^{\frac{w_N^{n+m}}{N}x}& = w_N^{0\cdot k}e^{\frac{w_N^{m}}{N}x}+...+w_N^{(m-1)\cdot k}e^{\frac{w_N^{2m-1}}{N}x}+w_N^{m\cdot k}e^{\frac{w_N^{2m}}{N}x}+...+w_N^{(N-1)\cdot k}e^{\frac{w_N^{(N-1)+m}x}{N}}\\
& = w_N^{0\cdot k}e^{\frac{w_N^{m}}{N}x}+...+w_N^{(m-1)\cdot k}e^{\frac{w_N^{N-1}}{N}x}+w_N^{m\cdot k}e^{\frac{w_N^{0}}{N}x}+...+w_N^{(N-1)\cdot k}e^{\frac{w_N^{(N-1)+m}x}{N}}\\
& = w_N^{mk}\left[w_N^{m\cdot k}e^{\frac{w_N^{m}}{N}x}+...+w_N^{(N-1)\cdot k}e^{\frac{w_N^{N-1}}{N}x}+w_N^{0\cdot k}e^{\frac{w_N^{0}}{N}x}+...+w_N^{(m-1)\cdot k}e^{\frac{w_N^{m-1}x}{N}}\right]\\
& = w_N^{mk}\left[w_N^{0\cdot k}e^{\frac{w_N^{0}}{N}x}+...+w_N^{(m-1)\cdot k}e^{\frac{w_N^{m-1}x}{N}}+w_N^{m\cdot k}e^{\frac{w_N^{m}}{N}x}+...+w_N^{(N-1)\cdot k}e^{\frac{w_N^{N-1}}{N}x}\right]\\
& = w_N^{mk}\sum_{n=0}^{N-1}w_N^{-nk}e^{\frac{w_N^{n}}{N}x}\\
& = \sum_{n=0}^{N-1}w_N^{-nk}w_N^{mk}e^{\frac{w_N^{n}}{N}x}
\end{align}
And as you mentioned, you can simplify this even further since $w_N^m=-1$ you get 
$$N^{k-1}k!\sum_{n=0}^{N-1}w_N^{-nk}e^{\frac{w_N^{n+m}}{N}x}=N^{k-1}k!(-1)^k\sum_{n=0}^{N-1}w_N^{-nk}e^{\frac{w_N^{n}}{N}x}$$
