Find an arbitrary power of a lower triangular matrix of size $3\times 3$ 
Let $F$ be a field and let $A=\begin{bmatrix}a&0&0\\1&a&0\\0&1&a\end{bmatrix}\in\mathscr{M}_{3\times 3}(F)$. Show that $$A^k=\begin{bmatrix}a^k&0&0\\ka^{k-1}&a^k&0\\\dfrac{1}{2}k(k-1)a^{k-2}&ka^{k-1}&a^k\end{bmatrix}$$ for all $k > 0$. (Exercise 797 from Golan, The Linear Algebra a Beginning Graduate Student Ought to Know.)

I know it can be proved by induction, but since the topic is about Krylov Spaces, eigenvalues and Jordan canonical form, I wonder if there is another way to solve this problem.

For a route different to induction, express the matrix as
$$A=aI_3+G$$
where $I_3$ is the identity matrix and $G$ is given by $$G=\begin{bmatrix}0&0&0\\1&0&0\\0&1&0\end{bmatrix}$$
A binomial expansion will yield
$$\begin{align}A^k&=(aI_3+G)^k\\&={k \choose 0}a^kI_3+{k \choose 1}a^{k-1}I_3G+{k \choose 2}a^{k-2}I_3G^2+\sum_{m=3}^k{k \choose m}a^{k-m}I_3G^m\\&=a^kI_3+ka^{k-1}I_3G+\frac{k(k-1)}{2}a^{k-2}I_3G^2+\sum_{m=3}^k{k \choose m}a^{k-m}I_3G^m\end{align}$$
Now $I_3G$ is given by
$$G=\begin{bmatrix}0&0&0\\1&0&0\\0&1&0\end{bmatrix}$$
while $I_3G^2$ is
$$G=\begin{bmatrix}0&0&0\\0&0&0\\1&0&0\end{bmatrix}$$
and $I_3G^m$ for $m\geq3$ is the zero matrix.
Using these results, we arrive at the given expression for $A^k$.
