Matrix demonstration $A^k$ Given a matrix $A = \begin{bmatrix}     7 & 4\\     -9 & -5 \end{bmatrix}$ $\in \mathcal{M2\times2}\, (\mathbb{R}) $
Show that $A^k = \begin{bmatrix}     1+6k & 4k\\     -9k & 1-6k \end{bmatrix} $
for every $k \in \mathbb{N}$
 A: Try a simple induction on $k$; it is clear that for $k = 1$,
$A^1 = A = \begin{bmatrix} 7 & 4 \\ -9 & -5 \end{bmatrix} = \begin{bmatrix} 1 + 6 \cdot 1 & 4 \cdot 1 \\ -9 \cdot 1 & 1 - 6 \cdot 1 \end{bmatrix}; \tag 1$
then assuming that for some $k$
$A^k = \begin{bmatrix} 1 + 6 \cdot k & 4 \cdot k \\ -9 \cdot k & 1 - 6 \cdot k \end{bmatrix}, \tag 2$
we find
$A^{k + 1} = A^kA = \begin{bmatrix} 1 + 6 \cdot k & 4 \cdot k \\ -9 \cdot k & 1 - 6 \cdot k \end{bmatrix}\begin{bmatrix} 7 & 4 \\ -9 & -5 \end{bmatrix} = \begin{bmatrix} 7 + 42k - 36k & 4 + 24k - 20k \\ -63k - 9 + 54k & -36k  - 5 + 30k \end{bmatrix}$
$= \begin{bmatrix} 7 + 6k & 4 + 4k \\ -9 -9k & - 5 -6k \end{bmatrix} = \begin{bmatrix} 1 + 6(k + 1) & 4(k + 1) \\ -9(k + 1) & 1 - 6(k + 1) \end{bmatrix}, \tag 3$
which is simply (2) with $k$ replaced by $k + 1$; we thus infer the formula (2) holds for all $k \ge 1$, as desired. $OE\Delta$.
A: Let $N = A-I$ and observe that $N^2=0$. Since $N$ and $I$ commute, we can expand via the binomial theorem: $$A^k = (I+N)^k = I+\binom k1N+\binom k2N^2+\dots = I+kN.$$ 
How did I hit upon this decomposition? I computed that $1$ was the only eigenvalue of $A$, but the only diagonalizable $2\times2$ matrices with repeated eigenvalues are multiples of the identity, so $A$ splits into the sum of the identity and a nilpotent matrix. That is, $$A = P\begin{bmatrix}1&1\\0&1\end{bmatrix}P^{-1} = I + P\begin{bmatrix}0&1\\0&0\end{bmatrix}P^{-1}$$ for some invertible matrix $P$.
