Nth power of the following matrix The matrix is 
$ \begin{pmatrix}0&1\\-1&0\end{pmatrix}^n $ 
for  $n=2 \implies \left(\begin{matrix}-1 & 0\\ 0 &-1\end{matrix}\right)$ for $n=3 \implies \begin{pmatrix}1&0\\0&1\end{pmatrix}$ for $n=4 \implies \begin{pmatrix}-1&0\\0&-1\end{pmatrix}$
so I assume that for every $n=2k $, where k is a natural number and bigger than $0$ the matrix will be 
$\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$ and for every $n=2k+1$ where k is a natural number and bigger than $0 $the matrix will be $\begin{pmatrix}1&0\\0&1\end{pmatrix}$
How can I prove it? probably with induction and how can I get easily the inverses of the matrices?

You just have to calculate the first power:
$$A=\begin{pmatrix}0&\!-1\\1&0\end{pmatrix}\;,\;\;A^2=\begin{pmatrix}\!-1&0\\0&-1\end{pmatrix}=-I$$
and that's all we need, since then
$$A^3=A\cdot A^2=-A\;,\;\;A^4=I\;,\;\;A^5=A;,\ldots\text{etc.}$$
