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

Consider the polynomial $$P(z)=-{_nC_1} z^{n-1}+{_nC_2}z^{n-2}-\cdots +(-1)^n.\tag{1}\label{1}$$ The answer shows that the numbers $y_i=\frac{1}{1-\alpha_i}$ are all roots of $P$, so they must be all the roots. Thus we must have $$P(z)=-{_nC_1}(z-y_1)\cdots(z-y_{n-1})\tag{2}\label{2}$$ Then comparing the coefficients of $z^{n-2}$ in $P$ from \eqref{1} and \eqref{2}, we find $${_nC_2}={_nC_1}\left(\sum_i y_i\right)$$ and thus $$\sum_i y_i=\frac{_nC_2}{_nC_1}.$$

Since you already obtained some clarification on the textbook's solution, I am introducing a different solution. You can also prove by noting that the roots of $x^n-1=0$ are $1,\xi,\xi^2,\ldots,\xi^{n-1}$, where $$\xi=e^{\frac{2\pi i}{n}}.$$ Therefore, we may take $a_k$ to be $\xi^k$ for $k=1,2,\ldots,n-1$. Now, $$\frac{1}{1-a_k}+\frac{1}{1-a_{n-k}}=\frac{1}{1-\xi^k}+\frac{1}{1-\xi^{n-k}}=\frac{1}{1-\xi^k}+\frac{1}{1-\frac{\xi^n}{\xi^k}}.$$ Since $\xi^n=1$, we obtain $$\frac{1}{1-a_k}+\frac{1}{1-a_{n-k}}=\frac{1}{1-\xi^k}+\frac{1}{1-\frac{1}{\xi^k}}=\frac{1}{1-\xi^k}+\frac{\xi^k}{\xi^k-1}=1.$$ Therefore, $$2\sum_{k=1}^{n-1}\frac{1}{1-a_k}=\sum_{k=1}^{n-1}\left(\frac{1}{1-a_k}+\frac{1}{1-a_{n-k}}\right)=\sum_{k=1}^{n-1}1=n-1,$$ so $$\sum_{k=1}^{n-1}\frac{1}{1-a_k}=\frac{n-1}{2}.$$