I am attempting to prove that the discriminant of some polynomial $f$ of degree $n$ with roots $\alpha_1, \dots, \alpha_n$ is given by $$ \Delta_f = (-1)^{\frac{n(n-1)}{2}} \prod\limits_{k = 1}^n f'(\alpha_k) $$
I found a solution in the first part of problem 1 here, but I am struggling to understand some of the logic. In particular, they state that $$ \frac{f'(x)}{f(x)} = \sum\limits_{i=1}^n \frac{1}{x - \alpha_i} \implies f'(\alpha_j) = \prod_{\substack{i = 1 \\ i \neq j}}^n (a_j - a_i). $$
I understand how they derive the initial summation, but I fail to see how the product follows. Any help would be appreciated!
What appears below assumes the leading coefficient is $1.$ \begin{align} \frac{f'(x)}{f(x)} & = \sum_{i=1}^n \frac 1 {x-\alpha_i}. \\[10pt] \text{Therefore } f'(x) & = f(x) \sum_{i=1}^n \frac 1 {x-\alpha_i} \\[10pt] & = (x-\alpha_1)\cdots(x-\alpha_n)\sum_{i=1}^n \frac 1 {x-\alpha_i} \\[10pt] & = \sum_{i=1}^n \frac{(x-\alpha_1)\cdots(x-\alpha_n)}{x-\alpha_i} \\[10pt] & = \overbrace{(x-\alpha_2)(x-\alpha_3)\cdots(x-\alpha_n)}^{\Large\text{omitting }\alpha_1} \, + \, \overbrace{(x-\alpha_1)(x-\alpha_3) \cdots(x-\alpha_j)}^{\Large\text{omitting }\alpha_2} \,+\, \cdots\cdots \\[10pt] \text{So } f'(\alpha_1) & = (\alpha_1-\alpha_2)(\alpha_1-\alpha_3)\cdots(\alpha_1-\alpha_n) + 0 + 0 + \cdots\cdots + 0. \end{align}