Find sum $u_0 u_1 + u_1u_2+...+u_{n-2}u_{n-1} $ I have 
$$ u_k = \cos\frac{2k\pi}{n} + i \sin\frac{2k\pi}{n}$$
And I should calculate:
$$ u_0 u_1 + u_1u_2+...+u_{n-2}u_{n-1}+u_{n-1}u_0 $$
But I have stucked:

Firstly I calculate 
$$u_0 u_1 + u_1u_2+...+u_{n-2}u_{n-1} $$
and put
$$ \alpha_k = u_k \cdot u_{k+1} = ... = e^{\frac{i\pi(2k+1)}{n}} $$
and sum
$$ \alpha_0 + ... + \alpha_{n-2} = ... = e^{\frac{i\pi}{n}} \cdot \frac{1-e^{\frac{2(n-1)i\pi}{n}}}{1-e^{\frac{2i\pi}{n}}}$$ and I don't know how ti finish that. If it comes to 
$$u_{n-1}u_0  =  e^{i\pi} = -1 $$

Note that $$\alpha_k = u_{k}\cdot u_{k+1} = e^{i\frac{2\pi}{n}(2k+1)} = e^{i\frac{2\pi}{n}}\cdot e^{i\frac{4\pi}{n}k}.$$
Therefore,
$$\sum_{k=0}^{n-2}{\alpha_k} = e^{i\frac{2\pi}{n}}\sum_{k=0}^{n-2}{\left(e^{i\frac{4\pi}{n}}\right)^{k}} = e^{i\frac{2\pi}{n}} \frac{1-e^{i\frac{4\pi(n-1)}{n}}}{1-e^{i\frac{4\pi}{n}}} = e^{i\frac{2\pi}{n}}\frac{1-e^{-i\frac{4\pi}{n}}}{1-e^{i\frac{4\pi}{n}}} = \frac{e^{i2\pi/n}-e^{-i2\pi/n}}{e^{i2\pi/n}(e^{-i2\pi/n}-e^{i2\pi/n})}=-e^{-i\frac{2\pi}{n}}.$$
