$A^n = I$ only if $n\equiv 0 \pmod{4}$ Is the following proof correct? In addition could you pleases suggest a more cleaner or shorter proof.
Theorem. Given any $n\in\mathbb{Z^+}$ then 
$$\begin{pmatrix}
0&0&-1\\
0&1&0\\
1&0&0\\
\end{pmatrix}^n
=
\begin{pmatrix}
1&0&0\\
0&1&0\\
0&0&1\\
\end{pmatrix}
$$
only if $4 \mid n$.
Proof. Assume that $n\in\mathbb{Z^+}$ and let $A$ denote the matrix 
$$
\begin{pmatrix}
0&0&-1\\
0&1&0\\
1&0&0\\
\end{pmatrix}
$$
We prove the contrapositive. Given the standard choice of basis for $\mathbb{R}^3$ we can see $A$ represents a $90^{\circ}$ clockwise rotation about the $y$-axis consequently we may write out $A^2$, $A^3$ and $A^4$ as follows:
$$
A^2 = 
\begin{pmatrix}
-1&0&0\\
0&1&0\\
0&0&-1\\
\end{pmatrix}
$$
$$
A^3 = 
\begin{pmatrix}
0&0&1\\
0&1&0\\
-1&0&0\\
\end{pmatrix}
$$
$$
A^4 = 
\begin{pmatrix}
1&0&0\\
0&1&0\\
0&0&1\\
\end{pmatrix}
$$
Now assume that $4\not\mid n$ then the division theorem implies that $n$ is of the form $4q+1$, $4q+2$ or $4q+3$. Considering the case where $4q+1$ we can see that
$$A^{4q+1} = (A^{4})^q\cdot A = I^q\cdot A = I\cdot A\neq I.$$
By similar reasoning we can make the same conclusion in the other two cases.
$\blacksquare$

Yes, it is correct, but I would not use a geometric argument here, since a very simple computation is enough to obtain $A^n$ for $n\in\{2,3,4\}$.
