$\frac{1}{z^2}$ is holomorphic 
I have to show that $z\mapsto\frac1{z^2}$ is holomorpic on $\mathbb C\setminus\{0\}$ and compute its $n$-th derivative 

I know that $\frac{1}{z^2}=\sum\limits_{n\ge0}(-1)^n(n+1)(z-1)^n$, so it has a power series representation. Is that sufficient to conclude that the function is analytic, which is stronger than holomorphic. I thought analytic means inifinitely differentiable and holomorphic only continuously differentiable, how can one then compute $n$-th derivative of a holomorphic function, if $n$ is greater than $1$? However I computed it;
$\left(\frac{1}{z^2}\right)^{(n)}=\sum\limits_{k\ge n}(-1)^k(n+1)n!(z-1)^{k-n}$ is that correct ?
 A: If we know that $f(z)$ and $g(z)$ have complex derivatives at $z=z_0$, then
$$
\begin{align}
&\lim_{z\to z_0}\frac{f(z)g(z)-f(z_0)g(z_0)}{z-z_0}\\
&=\lim_{z\to z_0}\frac{f(z)g(z)-f(z)g(z_0)+f(z)g(z_0)-f(z_0)g(z_0)}{z-z_0}\\
&=\lim_{z\to z_0}\frac{f(z)g(z)-f(z)g(z_0)}{z-z_0}
+\lim_{z\to z_0}\frac{f(z)g(z_0)-f(z_0)g(z_0)}{z-z_0}\\
&=\lim_{z\to z_0}f(z)\frac{g(z)-g(z_0)}{z-z_0}
+g(z_0)\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}\\[6pt]
&=f(z_0)g'(z_0)+f'(z_0)g(z_0)\tag{1}
\end{align}
$$
Using $(1)$, $\frac{\mathrm{d}}{\mathrm{d}z}1=\lim\limits_{z\to z_0}\frac{1-1}{z-z_0}=0$, $\frac{\mathrm{d}}{\mathrm{d}z}z=\lim\limits_{z\to z_0}\frac{z-z_0}{z-z_0}=1$,  and induction, we get for $n\ge0$
$$
\frac{\mathrm{d}}{\mathrm{d}z}z^n=nz^{n-1}\tag{2}
$$
If we know that $f(z)$ has a complex derivative at $z=z_0$ and that $f(z_0)\ne0$, then
$$
\begin{align}
\lim_{z\to z_0}\frac{\frac1{f(z)}-\frac1{f(z_0)}}{z-z_0}
&=\lim_{z\to z_0}\left[-\frac1{f(z)f(z_0)}\frac{f(z)-f(z_0)}{z-z_0}\right]\\
&=-\frac1{f(z_0)^2}f'(z_0)\tag{3}
\end{align}
$$
Thus, $1/f(z)$ also has a complex derivative at $z=z_0$.
Using $(2)$ and $(3)$, we see that we can extend $(2)$ to $n\in\mathbb{Z}$:
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}z}z^{-n}
&=-\frac1{z^{2n}}nz^{n-1}\\
&=-nz^{-n-1}\tag{4}
\end{align}
$$
Equation $(4)$ for $n=2$, says that $z^{-2}$ has a complex derivative for $z\ne0$; therefore, it is holomorphic on $\mathbb{C}\setminus\{0\}$.
Furthermore, by induction on $(4)$, we get
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}z}z^{-2}&=-2z^{-3}\\
\frac{\mathrm{d}^2}{\mathrm{d}z^2}z^{-2}&=6z^{-4}\\
\frac{\mathrm{d}^3}{\mathrm{d}z^2}z^{-2}&=-24z^{-5}\\
&\vdots\\
\frac{\mathrm{d}^n}{\mathrm{d}z^n}z^{-2}&=(-1)^n(n+1)!\,z^{-n-2}\tag{5}
\end{align}
$$
A: Given$$f(z) = {z^{ - 2}}$$
then, for$$z = x + iy$$
we may write$$f(x + iy) = \frac{{{x^2} - {y^2}}}{{{{({x^2} + {y^2})}^2}}} + i\frac{{ - 2xy}}{{{{({x^2} + {y^2})}^2}}}$$
For$$\begin{gathered}
  u(x,y) = \frac{{{x^2} - {y^2}}}{{{{({x^2} + {y^2})}^2}}} \hfill \\
  v(x,y) = \frac{{ - 2xy}}{{{{({x^2} + {y^2})}^2}}} \hfill \\ 
\end{gathered}$$
Cauchy-Riemann Equations are satisfied.
For n-th order derivative, we get for $n \in \mathbb{N}$:
$$\frac{{{d^n}f}}{{d{z^n}}}(z) = \left\{ {\begin{array}{*{20}{c}}
  {{{( - 1)}^{2n - 1}}(2n)! \cdot z \cdot f{{(z)}^{2n - 1}}} \\ 
  {{{( - 1)}^{2n}}(2n + 1)!f{{(z)}^{2n - 1}}} 
\end{array}} \right.$$
One gets this by carefully calculating:
$$\begin{gathered}
  \frac{{df}}{{d{z^1}}}(z) = {( - 1)^1} \cdot 2! \cdot z \cdot f{(z)^2} \hfill \\
  \frac{{{d^3}f}}{{d{z^3}}}(z) = {( - 1)^3} \cdot 4! \cdot z \cdot f{(z)^3} \hfill \\
  \frac{{{d^5}f}}{{d{z^5}}}(z) = {( - 1)^5} \cdot 6! \cdot z \cdot f{(z)^4} \hfill \\
   \vdots  \hfill \\
   \hfill \\
  \frac{{{d^2}f}}{{d{z^2}}}(z) = 3! \cdot f{(z)^2} \hfill \\
  \frac{{{d^4}f}}{{d{z^4}}}(z) = 5! \cdot f{(z)^3} \hfill \\
  \frac{{{d^6}f}}{{d{z^6}}}(z) = 7! \cdot f{(z)^4} \hfill \\
   \vdots  \hfill \\ 
\end{gathered}$$
