Why does $\sum_{n=0}^k \cos^{2k}\left(x + \frac{n \pi}{k+1}\right) = \frac{(k+1)\cdot(2k)!}{2^{2k} \cdot k!^2}$? In the paper "A Parametric Texture Model based on Joint Statistics of Complex Wavelet Coefficients", the authors use this equation for the angular part of the filter in polar coordinates:
$$\sum_{n=0}^k \cos^{2k}\left(x + \frac{n \pi}{k+1}\right)$$
My friend and I have tested many values of $k > 1$, and in each case this summation is equal to 
$$\frac{(k+1)\cdot(2k)!}{2^{2k} \cdot k!^2}$$
The paper asserts this as well.
We are interested in having an analytic explanation of this equality, if it really holds. How can we derive this algebraically? 
TL;DR

Is this true, and if so, why?
  $$\sum_{n=0}^k \cos^{2k}\left(x + \frac{n \pi}{k+1}\right) = \frac{(k+1)\cdot(2k)!}{2^{2k} \cdot k!^2}$$


First note that
$$
\left[\cos\left(x+\frac{n\pi}{k+1}\right)\right]^{2k} = \frac{1}{2^{2k}} \exp\left(-i2k\left(x+\frac{n\pi}{k+1}\right)\right)\sum_{j=0}^{2k} {2k \choose j} \exp\left(i2j\left(x+\frac{n\pi}{k+1}\right)\right)
$$
Then,
\begin{align}
\sum_{n=0}^k \left[\cos\left(x+\frac{n\pi}{k+1}\right)\right]^{2k} &= \frac{1}{2^{2k}} \sum_{n=0}^k \exp\left(-i2k\left(x+\frac{n\pi}{k+1}\right)\right)\sum_{j=0}^{2k} {2k \choose j} \exp\left(i2j\left(x+\frac{n\pi}{k+1}\right)\right) \\
&=  \frac{1}{2^{2k}} \sum_{j=0}^{2k} {2k \choose j} \sum_{n=0}^k  \exp\left(i2\left(x+\frac{n\pi}{k+1}\right)(j-k)\right) \tag{1}
\end{align}
Now, suppose that $j \neq k$. Then,
\begin{align}
(1) &= \frac{1}{2^{2k}} \sum_{j=0}^{2k} {2k \choose j} \exp(i2x(j-k)) \sum_{n=0}^k  \left[\exp\left(i2\frac{\pi}{k+1}(j-k)\right)\right]^n \\
&= \frac{1}{2^{2k}} \sum_{j=0}^{2k} {2k \choose j} \exp(i2x(j-k)) \frac{1 - \left[\exp\left(i2\frac{\pi}{k+1}(j-k)\right)\right]^{k+1}}{1-\exp\left(i2\frac{\pi}{k+1}(j-k)\right)} \\
&= 0
\end{align}
Thus, considering $(1)$ only when $j=k$, we have the result.
