Powers of complex permutation matrix I am looking for $P^{100}$, where
$$
P=
        \begin{bmatrix}
        0 & i & 0 \\
        0 & 0 & i \\
        i & 0 & 0 \\
        \end{bmatrix}.
$$
The cycle repeats for $P^{13} = P$ and I find $P^{100}=P^{4}$. Is there a way to arrive at this answer quicker than by computing $P^2, P^3, \dots P^{13}$?

The characteristic polynomial of this matrix is $-x^3 - i$. We know (Cayley-Hamilton) that $P$ satisfies $-P^3 - iI = 0$, so $P^3 = -iI$ and $P^{12} = (-i)^4I = I$.
