Matrix raised to 14th power Calculate $\left(\begin{matrix} 6&1&0\\0&6&1\\0&0&6\end{matrix}\right)^{14}$
Whould I do it one by one, and then find a pattern? I sense $6^{14}$ on the diagonal, and zeroes in the "lower triangle", but the "upper triangle" I'm not sure. Was thinking $14 \cdot 6^{13} $ but that's not correct.

I didn't notice that this was suggested by Harald Hanche-Olsen until just now. Consider this an expansion on his answer.
Since the identity matrix commutes with any matrix, we can use the binomial theorem with
$$
\left(6\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}+\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}\right)^n
$$
while noting that
$$
\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}^2=\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}
\quad\text{and}\quad
\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}^3=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}
$$
To get
$$
\begin{bmatrix}6&1&0\\0&6&1\\0&0&6\end{bmatrix}^n
=6^n\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}+6^{n-1}\binom{n}{1}\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}+6^{n-2}\binom{n}{2}\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}
$$
