<s>[INST] Find $A^n$ for all $n \in \mathbb{N}$. I'm learning linear algebra and need help with the following problem:

Let $A = \begin{pmatrix}-2 & -3 & -3\\-1 & 0 & -1\\5 & 5 & 6\end{pmatrix} \in M_{3x3}(\mathbb{R}).$ Find $A^n$ for all $n \in \mathbb{N}$.

My first thought was to compute $A^2$, $A^3$, $A^4$ and see if a pattern emerge. I used Mathematica to compute the power of $A$ to save me some time. The computations gave
$A^2 = \begin{pmatrix}-8 & -9 & -9\\-3 & -2 & -3\\15 & 15 & 16\end{pmatrix}, \quad A^3 = \begin{pmatrix}-20 & -21 & -21\\-7 & -6 & -7\\35 & 35 & 36\end{pmatrix}, \quad A^4 = \begin{pmatrix}-44 & -45 & -45\\-15 & -14 & -15\\75 & 75 & 76\end{pmatrix}$.
The above results are clearly showing a pattern relating the first four powers of $A$, for instance $(a_{31}) = (a_{32}) = (a_{33}) + 1$, but I'm unable to write the general form of a matrix $A^{k}$. 
$(Q1)$ How do I find/write the general form of a matrix $A^{k}$? 
$(Q2)$ I only compute the first four powers of $A$. How do I know that this pattern won't fail for some integer $k$? 
 [/INST] The matrix is similar to the diagonal matrix $D=diag(1,1,2)$. Hence we have
$$
A^n=(SDS^{-1})^n=SD^nS^{-1}
$$
with 
$$
S=\begin{pmatrix} -5 & -4 & 3\cr -1 & -1 & 1\cr 6 & 5 & -5\end{pmatrix}.
$$
</s>