if 1, $\alpha_1$, $\alpha_2$, $\alpha_3$, $\ldots$, $\alpha_{n-1}$ are nth roots of unity then... 
if 1, $\alpha_1$, $\alpha_2$, $\alpha_3$, $\ldots$, $\alpha_{n-1}$ are nth roots of unity then
  $$\frac{1}{1-\alpha_1} + \frac{1}{1-\alpha_2} + \frac{1}{1-\alpha_3}+\ldots+\frac{1}{1-\alpha_n} = ?$$

Now this has the solution too, but I do not understand the last step of the solution so here is the solution from the book.
$1,\, \alpha_1, \, \alpha_2,\, \ldots, \, \alpha_n$ are the $n^\text{th}$ of unity. These are the roots of $x^n-1=0$
Let $y=\frac{1}{1-\alpha}$ where $\alpha = \alpha_1, \, \alpha_2, \, \alpha_3, \, \ldots,  \, \alpha_n$ 
$$1 - \alpha = \frac{1}{y} \Rightarrow \alpha = \frac{y-1}{y}.$$
But $\alpha$ is a root of $x^n-1=0 \therefore \alpha^n=1 \Rightarrow (y-1)^n = y^n$ 
$$\Rightarrow y^n - _nC_1y^{n-1} + _nC_2y^{n-2}-\ldots+(-1)^n = y^n \\
\Rightarrow  - _nC_1y^{n-1} + _nC_2y^{n-2}-\ldots+(-1)^n = 0.$$
Sum of roots 
$$\frac{1}{1-\alpha_1}+\frac{1}{1-\alpha_2}+\ldots+\frac{1}{1-\alpha_{n-1}} = \frac{_nC_2}{_nC_1} = \frac{n-1}{2}$$
So this last part from "Sum of roots" I do not understand. I cannot see how this last shape relates to this binomial theorem notation. Can anyone help?

The sum of the roots of $a_nx^{n}+a_{n-1}x^{n-1}+\cdots +a_0$ is $-\frac {a_{n-1}} {a_n}$.
