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?

$${{\left( \begin{matrix}
   \cos \,\phi  & -\sin \,\phi   \\
   \sin \,\phi  & \cos \,\phi   \\
\end{matrix} \right)}^{n}}=\left( \begin{matrix}
   \cos \,n\phi  & -\sin \ n\phi   \\
   \sin \,n\phi  & \cos \,n\phi   \\
\end{matrix} \right)$$
We know $A=\begin{pmatrix}0&\!-1\\1&0\end{pmatrix}$
then set $\phi=\frac{\pi}{2}$.
