In my question I am referring to the proof, that can be found here.
There $P_j(x)$ is defined as : $P_j(x) = \frac{f(x)}{(x-x_j)f'(x_j)}$, where $f(x) = (x-x_1)\dots(x-x_n)$
Then, supposedly, $P(x) = 1 $ if $x \in B:= \{x_1 , \dots, x_n \}$ - but how/why?
Using this, I would get: $$1 = P(x_1) = \frac{(x_1-x_2)\dots(x_1-x_n)}{f'(x_1)} \Rightarrow f'(x_1) = (x_1-x_2)\dots(x_1-x_n) $$
which looks to me like some sort of basic identity.
I also tried looking at $f'(x)$ in general as: $f'(x) = x' (x-x_2)\dots(x-x_n) - x_1 ((x-x_2)\dots(x-x_n))' \overset{\mbox{?}}{\Rightarrow} x_1 (x-x_2)\dots(x-x_n) = 0 $ if $x = x_1$
I don't really see how this steps follow, (if at all) - thus I would be happy about any constructive comment/answer regarding the construction of the $P_j(x)$ and the role $f'(x)$ has in it, as I find that the confusing part.
You have, \begin{align} f'(x)&=(x-x_1)'(x-x_2)\cdots(x-x_n)+(x-x_1)\,[(x-x_2)\cdots(x-x_n)]'\\ \ \\ &=(x-x_2)\cdots(x-x_n)+(x-x_1)\,[(x-x_2)\cdots(x-x_n)]'. \end{align} When you evaluate at $x_1$, the second term is zero, so $$ f'(x_1)=(x_1-x_2)(x_1-x_3)\cdots(x_1-x_n). $$