Proving $\frac{1}{1-x} \circ \frac{1}{1-x} = 1 - \frac{1}{x}$ from a series point of view

For your problem with power series for $\, f(x) := 1/(1-x), \,\, g(x) := 1-1/x, \,$ you would like to center each power series about the same number when the functions are composed. In this case the common center is $\, \omega, \,$ a primitive sixth root of unity because $\, \omega = f(\omega) = g(\omega) \,$ is a fixed point of both $\,f\,$ and $\,g.\,$ Thus, let $\, y := x-\omega \,$ be the local variable. Check that the two power series expansions in powers of $\, y\,$ are $$ f(x) = \omega + \omega^2 y - y^2 - \omega y^3 -\omega^2 y^4 + y^5 + O(y^6) \, = \frac{\omega + \omega^2 y - y^2} {(1 + y^3)}, \tag1$$ $$ g(x) = \omega - \omega y + y^2 + \omega^2 y^3 - \omega y^4 + y^5 + O(y^6) \, = \omega + \frac{y}{\omega(\omega+y)}. \tag2$$ The radius of convergence for both series is $\,1\,$ centered at $\, \omega \,$ and includes $\, 0<x<1. \,$ Check with composition the following equations: $$ f(f(x)) = g(x), \,\, f(g(x)) = x, \,\, g(f(x)) = x, \,\, g(g(x)) = f(x). \tag3$$