Determine all positive powers. Determine all positive powers of the $5 \times 5$ matrix...
$$C=\begin{bmatrix}
0&0&0&0&0 \\
1&0&0&0&0 \\
0&1&0&0&0 \\
0&0&1&0&0 \\
0&0&0&1&0
\end{bmatrix}$$
I know that taking powers of $C$ just shifts each $1$ down a row, replacing each $1$ with a $0$. The right most $1$ is shifted out of the matrix and replaced with a $0$. This happens for $C^2$, $C^3$, $C^4$, and $C^5$, where the sub-diagonal of $1's$ is shifted down until $C^5$ at which time the matrix becomes the $5 \times 5$ zero matrix.
This is the solution I came up with...
\begin{align}
C^2 &= \begin{bmatrix}
0&0&0&0&0 \\
0&0&0&0&0 \\
1&0&0&0&0 \\
0&1&0&0&0 \\
0&0&1&0&0
\end{bmatrix},  \\
C^3 &= \begin{bmatrix}
0&0&0&0&0 \\
0&0&0&0&0 \\
0&0&0&0&0 \\
1&0&0&0&0 \\
0&1&0&0&0
\end{bmatrix},  \\
C^4 &= \begin{bmatrix}
0&0&0&0&0 \\
0&0&0&0&0 \\
0&0&0&0&0 \\
0&0&0&0&0 \\
1&0&0&0&0
\end{bmatrix},  \\
C^k &= \begin{bmatrix}
0&0&0&0&0 \\
0&0&0&0&0 \\
0&0&0&0&0 \\
0&0&0&0&0 \\
0&0&0&0&0
\end{bmatrix}, k\geq5
\end{align}
Is there a more general way to express $C^2$, $C^3$, and $C^4$ without explicitly writing them out? Or asked another way, is it possible to express $C^k$ for all positive powers without having to explicitly write out $C^2$, $C^3$, and $C^4$?

$C$ is the adjacency matrix of a directed graph $5 \to 4 \to 3 \to 2 \to 1$.  $(C^k)_{ij}$ is then the number of paths of length $k$ starting at $i$ and ending at $j$.  This is $1$ if $i-j = k$, otherwise $0$.  Of course if $k \ge 5$ all entries are $0$.
