<s>[INST] 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?
 [/INST] 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.
</s>