Find $k^{th}$ power of a square matrix I am trying to find the $A^{k}$, for all  $k \geq 2$  of a matrix, \begin{pmatrix}  a & b \\ 0 & 1  \end{pmatrix} 
My approach:
$A^{2}=\begin{pmatrix}  a^2 & ab+b \\ 0 & 1 \end{pmatrix}$
$A^{3}=\begin{pmatrix}  a^3 & a^{2}b+ab+b \\ 0 & 1 \end{pmatrix}$
$A^{4}=\begin{pmatrix}  a^4 & a^{3}b+a^{2}b+ab+b \\ 0 & 1 \end{pmatrix}$
$A^{5}=\begin{pmatrix}  a^5 & a^{4}b+a^{3}b+a^{2}b+ab+b  \\ 0 & 1 \end{pmatrix}$
Continuing this way, we obtain
$A^{k}=\begin{pmatrix}  a^k & (a^{k-2}+a^{k-3}+a^{k-4}+.....+1)b  \\ 0 & 1 \end{pmatrix}$ 
I am stuck here! I was wondering if you could give me some hints to move further. I appreciate your time. 

I'm partial to the SVD for this. The eigenvalues are $a$ and $1$ associated with eigenvectors $(1,0)$ and $(b,1-a)$ respectively.
So, we get
$$\begin{bmatrix}
a & b \\
0 & 1
\end{bmatrix}^n= \begin{bmatrix}
1 & b \\
0 & 1-a
\end{bmatrix} \begin{bmatrix}
a^n & 0 \\
0 & 1
\end{bmatrix} \begin{bmatrix}
1 & \frac{-b}{1-a} \\
0 & \frac{1}{1-a}
\end{bmatrix} = \begin{bmatrix}
a^n & \frac{b(1-a^n)}{1-a} \\
0 & 1
\end{bmatrix}$$
