$\frac{1}{z^2}$ 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 ?

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:
$$\begin{gathered}
  \frac{{df}}{{d{z^1}}}(z) = {( - 1)^1} \cdot 2! \cdot z \cdot f{(z)^2} \hfill \\
  \frac{{{d^3}f}}{{d{z^3}}}(z) = {( - 1)^3} \cdot 4! \cdot z \cdot f{(z)^3} \hfill \\
  \frac{{{d^5}f}}{{d{z^5}}}(z) = {( - 1)^5} \cdot 6! \cdot z \cdot f{(z)^4} \hfill \\
   \vdots  \hfill \\
   \hfill \\
  \frac{{{d^2}f}}{{d{z^2}}}(z) = 3! \cdot f{(z)^2} \hfill \\
  \frac{{{d^4}f}}{{d{z^4}}}(z) = 5! \cdot f{(z)^3} \hfill \\
  \frac{{{d^6}f}}{{d{z^6}}}(z) = 7! \cdot f{(z)^4} \hfill \\
   \vdots  \hfill \\ 
\end{gathered}$$
