How can I show that $\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^n = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$? Well, the original task was to figure out what the following expression evaluates to for any $n$.
$$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^{\large n}$$
By trying out different values of $n$, I found the pattern:
$$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^{\large n} = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$$
But I have yet to figure out how to prove it algebraically. 
Suggestions?

Powers of matrices occur in solving recurrence relations.
If you write
$$
\begin{pmatrix} x_{n+1} \\ y_{n+1} \end{pmatrix} =
\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix} \begin{pmatrix} x_n \\ y_n \end{pmatrix}
$$
then clearly
$$
\begin{pmatrix} x_{n} \\ y_{n} \end{pmatrix} =
\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^n \begin{pmatrix} x_0 \\ y_0 \end{pmatrix}
$$
and also
$$
x_{n+1}=x_n+y_n, \qquad y_{n+1}=y_n
$$
from which you get
$$
x_n = x_0 + n\  y_0, \qquad y_n = y_0
$$
The first column of $A^n$ is given by taking $x_0=1$ and $y_0=0$, and so is $(1 \ 0)^T$.
The second column is given by taking $x_0=0$ and $y_0=1$, and so is $(n \ 1)^T$.
