Binomial expansion of $(1 - w z^{-1})^{-1}$ being $\sum_{l=0}^{\infty} w^l z^{-l}$ I am reading "Neural Networks and Learning Machines", Third Edition, Simon Haykin and at the beginning I came across at something that perplexes me.
I'll quote a whole page because some contexts may be important and I fail to notice them. But the question I am asking is: is it so that:
$$(1 - w z^{-1})^{-1} = \sum_{l=0}^{\infty} w^l z^{-l}$$
?
And now the quote (I won't use the quote sign ">" in order not to mess with  the equations):

Feedback is said to exist in a dynamic system whenever the output of an element in the
system influences in part the input applied to that particular element, thereby giving
rise to one or more closed paths for the transmission of signals around the system.
[...] Moreover, it plays a major role in the study of a special class of neural
networks known as recurrent networks. Figure 12 shows the signal-flow graph of a single-loop
feedback system, where the input signal $x_j(n)$, internal signal xj (n), and output
signal $y_k(n)$ are functions of the discrete-time variable n. The system is assumed to be
linear, consisting of a forward path and a feedback path that are characterized by the “operators” A and B, respectively. In particular, the output of the forward channel
determines in part its own output through the feedback channel. From Fig. 12, we readily
note the input–output relationships
$$y_k(n) = \textbf{A}[x'_j(n)]\qquad(16)$$
and
$$x'_j(n) = x_j(n) + \textbf{B}[y_k(n)]\quad(17)$$
where the square brackets are included to emphasize that A and B act as operators.
Eliminating $x_j(n)$ between Eqs. (16) and (17), we get
$$y_k(n) = \frac{\textbf{A}}{1-\textbf{AB}}[x_j(n)]\qquad(18)$$
We refer to A/(1 - AB) as the closed-loop operator of the system, and to AB as the
open-loop operator. In general, the open-loop operator is noncommutative in that BA $\neq$ AB.
Consider, for example, the single-loop feedback system shown in Fig. 13a, for which
A is a fixed weight w and B is a unit-delay operator $z^{-1}$, whose output is delayed with
respect to the input by one time unit.We may then express the closed-loop operator of
the system as
$$\frac{\textbf{A}}{1-\textbf{AB}} = \frac{w}{1-w z^{-1}}=w(1 - w z^{-1})^{-1}$$
Using the binomial expansion for $(1 - w z^{-1})^{-1}$, we may rewrite the closed-loop operator
of the system as
$$\frac{\textbf{A}}{1-\textbf{AB}} = w\sum_{l=0}^{\infty} w^l z^{-l}\qquad (19)$$

Fig 12
Fig 13a

I am studying the same book and pondered on the same. Here's the simple explanation:
$$(1 - w z^{-1})^{-1} = \sum_{l=0}^{\infty} w^l z^{-l}\qquad (19)\\\\$$
$$\text{Solving L.H.S}$$
$$\text{Let}  \qquad  w z^{-1} = m\\$$
$$\text{Thus,}\qquad (1 - w z^{-1})^{-1} = (1 - m)^{-1}      \qquad (20)\\$$
$$(1 - m)^{-1} = 1 + m + \frac{2{m^2}}{2} + ... \qquad = \sum_{l=0}^{\infty} m^l \qquad (21)$$
$$\text{Substituting the value of m in eqn. 21, we get R.H.S}$$
$$Hence Proved$$
