Finding $A^n$ for a matrix I have a matrix $$
A =
\left[ {\begin{array}{cc}
 1 & c  \\
 0 & d  \\
 \end{array} } \right]
$$
with $c$ and $d$ constant. I need to find $A^n$ ($n$ positive) and then need to prove that formula using induction.
I would like to check that the formula I derived is correct: 
$$
A^n =
\left[ {\begin{array}{cc}
 1 & c^{n-2}(dc + c)  \\
 0 & d^n  \\
 \end{array} } \right]
$$
If this is correct, how can I prove this? I suppose I can write $A^{n+1} = A^n A$, which would be
$$
\left[ {\begin{array}{cc}
 1 & c^{n-2}(dc + c)  \\
 0 & d^n  \\
 \end{array} } \right]
\left[ {\begin{array}{cc}
 1 & c  \\
 0 & d  \\
 \end{array} } \right]

$$
But then what would I do?
Thanks.

For this particular matrix, it is easy enough to compute the first few powers and conjecture a guess to prove by induction, but there are a lot of matrices (even $2\times 2$ matrices) where trying to find a pattern is significantly harder.  The following method works more generally:


*

*Diagonalize $A$, that is, write $A=SDS^{-1}$ where $S$ is invertible and $D$ is diagonal.  In general, you may have to use Jordan normal form, which will make things complicated but still solvable.

*$A^n=(SDS^{-1})^n= SD^nS^{-1}$.  This is easy to compute because if $D=\operatorname{diag}(d_1,d_2,\ldots, d_k)$, then $D^n=\operatorname{diag}(d^n_1,d^n_2,\ldots, d^n_k)$.
In our particular case, if $d\neq 1$, the matrix is diagonalizable, and we have $D=\pmatrix{1 & 0 \\ 0 & d}$, $S=\pmatrix{1 & \frac{c}{d-1} \\ 0 & 1}$, and so $$A^n=\pmatrix{1 & \frac{c}{d-1} \\ 0 & 1}\pmatrix{1 & 0 \\ 0 & d^n}\pmatrix{1 & \frac{-c}{d-1} \\ 0 & 1}$$
