I am studing the cauchy derivates theorem and there is one step which i don't understand. When we are proving for $f^{(n)}$ for math induction.
The hipothese are: $$f^{(n-1)}(z):=\frac{(n-1)!}{2 \pi i} \int_{\partial D} \frac{f(\xi)}{(\xi - z )^n}d \xi$$
The step is:
\begin{align} f^{(n)}&=\lim_{h \to 0} \frac{f^{(n-1)}(z+h)-f^{(n-1)}(z)}{h} \\&=\frac{(n-1)!}{2 \pi i} \lim_{h \to 0} \frac{1}{h} \bigg[ \int_{\partial D} \frac{f(\xi)}{(\xi - z - h )^n}d \xi - \int_{\partial D} \frac{f(\xi)}{(\xi - z )^n}d \xi\bigg] \\&=\frac{(n-1)!}{2 \pi i} \lim_{h \to 0} \frac{1}{h} \int_{\partial D} f(\xi)\frac{(\xi -z)^n-(\xi -z-h)^n}{(\xi - z - h )^n(\xi-z)^n}d \xi &(1) \\ &=\frac{(n-1)!}{2 \pi i} \lim_{h \to 0} \frac{1}{h} \int_{\partial D} f(\xi)\frac{nh(\xi -z)^{n-1}-\cdots}{(\xi - z - h )^n(\xi-z)^n}d \xi &(2) \\ &=\frac{(n-1)!}{2 \pi i} \int_{\partial D}\lim_{h \to 0} \bigg[ f(\xi)\frac{n-\cdots}{(\xi - z - h )^n(\xi-z)} \bigg]d \xi &(3) \\ &\qquad \qquad \qquad \qquad \qquad \qquad \qquad \vdots \end{align}
I don't understand the step from $(1)$ to $(2)$.
Expanding the second numerator fraction term using the Binomial theorem when going from $(1)$ to $(2)$ gives
$$\begin{equation}\begin{aligned} & (\xi - z)^n - \left((\xi - z) - h\right)^n \\ & = (\xi - z)^n - \left((\xi - z)^n - nh(\xi - z)^{n-1} + \ldots\right) \\ & = (\xi - z)^n - (\xi - z)^n + nh(\xi - z)^{n-1} - \ldots \\ & = nh(\xi - z)^{n-1} - \ldots \end{aligned}\end{equation}\tag{1}\label{eq1A}$$