Understanding a precursor to the complex Taylor Series

Question

First $f_n(z)$ is defined as

$$ f_n(z) = {f_{n-1(z) - f_{n-1}(a)} \over z - a} $$

for all $1, \ldots , n$.and s.t. $f_n(a) = f_{n-1}'(a)$. From this it is derived that

$$ f(z) = f(a) + (z-a)f_1(z) $$ $$ f_1(z) = f_1(a) + (z-a)f_2(z) $$ $$ \vdots $$ $$ f_{n-1}(z) = f_{n-1}(a) + (z-a)f_n(z) $$

so that evidently we have

$$ f(z) = f(a) + (z-a)f_1(a) + \underbrace{(z-a)^2f_2(a) + \ldots + (z-a)^{n-1}f_{n-1}(a)+(z-a)^n f_n(z)}_{\text{so that this must be $0$}} $$

Question: How exactly is this last equation being derived from the above ones?

Answer 1

Induction: Suppose that $$f(z) = f(a) + (z-a)f_1(a) + (z-a)^2f_2(a) + \ldots + (z-a)^{n-2}f_{n-2}(a)+(z-a)^{n-1} f_{n-1}(z).$$ Then since $$(z-a)^{n-1}f_{n-1}(z) = (z-a)^{n-1}f_{n-1}(a) + (z-a)^nf_n(z),$$ you get your expression.