Algebraic manipulation in Vaughan's book on the circle method On page 125 of Vaughan's1 book on the circle method we have in Chapter 5.2 on Waring's problem for $k=1$.
I do not understand how to get from $(1)$ to $(2)$:
\begin{align}
f(z)^s &= \frac{1}{(1-z)^s}\tag{1}\\
&=\frac{1}{(s-1)!} \frac {d^{s-1}} {dz^{s-1}} \frac 1{1-z}\tag{2}
\end{align}
I wonder what the step by step manipulations that go from $(1)$ to $(2)$?
[1]  R. C. Vaughan, The Hardy-Littlewood method Cambridge University Press (1997)
 A: It might make more sense to look at (2) first.  Here it is again with brackets to make it clear that it contains an ${(s-1)}$th-order derivative:
$$
\frac{1}{(s-1)!} \cdot \frac{d^{s-1}}{dz^{s-1}} \left[ \frac{1}{1-z} \right].
$$
So you want to differentiate the quantity in the brackets $s-1$ times.  Recall that
\begin{align}
\frac{d}{dz} \left[ \frac{(n-1)!}{(1-z)^n} \right] &= \\
&= (n-1)! \cdot \frac{d}{dz} \left[ \frac{1}{(1-z)^n} \right] \\ &= (n-1)! \cdot \frac{-n}{(1-z)^{n+1}} \cdot \frac{d}{dz} [1-z] \\ &= (n-1)! \cdot \frac{-n}{(1-z)^{n+1}} \cdot (-1) \\ &= \frac{n!}{(1-z)^{n+1}}.
\end{align}
Each time you differentiate, you are increasing the integer in the factorial by one (it starts at 0) and increasing the exponent in the denominator by one (it starts at 1).  So, at the end of $s-1$ derivatives, you are left with
$$
\frac{d^{s-1}}{dz^{s-1}} \left[ \frac{1}{1-z} \right] = \frac{(s-1)!}{(1-z)^s}.
$$
Plugging that expression back into (2) from the book, the $(s-1)!$ in the numerator cancels with the $(s-1)!$ in the denominator, yielding (1) from the book.
