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?

Although another answer is hardly needed at this point, here is yet another way of thinking about this. Note that, for any real numbers $x$ and $y$, one has
$$ \begin{pmatrix}
1 & x \\ 0 & 1 \\
\end{pmatrix} \cdot
\begin{pmatrix}
1 & y \\ 0 & 1 \\
\end{pmatrix} =
\begin{pmatrix}
1 & x+y \\ 0 & 1 \\
\end{pmatrix}.$$
We have learned that:


*

*The set of matrices of the form $\begin{pmatrix}
1 & \text{stuff} \\ 0 & 1 \\
\end{pmatrix}$ is closed under multiplication.

*To multiply two such matrices, you just need to add the stuff in the top right corner.


As a corollary, we conclude that:
$$ 
\begin{pmatrix}
1 & 1 \\ 0 & 1 \\
\end{pmatrix}^n
=
\underbrace{
\begin{pmatrix}
1 & 1 \\ 0 & 1 \\
\end{pmatrix}
\cdots
\begin{pmatrix}
1 & 1 \\ 0 & 1 \\
\end{pmatrix}
}_{n \text{ times}}
= 
\begin{pmatrix}
1 & \underbrace{1+ \ldots + 1}_{n \text{ times}} \\ 0 & 1 \\
\end{pmatrix}
=
\begin{pmatrix}
1 & n \\ 0 & 1 \\
\end{pmatrix}.
$$

Remark: If you like, you can think of the map
$$ x \mapsto \begin{pmatrix}
1 & x \\ 0 & 1 \\
\end{pmatrix}$$
as being a homomorphism from the real numbers under addition to the $2 \times 2$ invertible real matrices under multiplication.
