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.

Write the matrix as $6I+N$ where $$N=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}$$ and expand $(6I+N)^{14}$ using the binomial formula (which is valid here because $I$ and $N$ commute). Note that since $N^3=0$, you don't need to compute most of the coefficients.

Write it as $\left(\begin{matrix} 6&1&0\\0&6&1\\0&0&6\end{matrix}\right)=\left(\begin{matrix} 6&0&0\\0&6&0\\0&0&6\end{matrix}\right)+\left(\begin{matrix} 0&1&0\\0&0&1\\0&0&0\end{matrix}\right)$, use the binomial formula and see what you can say about powers of these two matrices.

$$A^n = \begin{bmatrix} 6^n & n \cdot 6^{n-1} & \dbinom{n}2 6^{n-2}\\ 0 & 6^n & n \cdot 6^{n-1}\\ 0 & 0 & 6^n\end{bmatrix}$$ Prove this by induction.

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 $$ \begin{bmatrix}6&1&0\\0&6&1\\0&0&6\end{bmatrix}^n =6^n\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}+6^{n-1}\binom{n}{1}\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}+6^{n-2}\binom{n}{2}\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix} $$