<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] $$
\left(%
\begin{array}{cc}
1 & 1
\\
0 & 1
\end{array}\right)
=
1 + A
\quad\mbox{where}\quad
A
=
\left(%
\begin{array}{cc}
0 & 1
\\
0 & 0
\end{array}\right)
$$
Notice that $A^{2} = 0$. It means that a function ${\rm f}\left(A\right)$ is
${\it linear}$ in $A$. For instance,
$\left(1 + \mu A\right)^{n} = \alpha + \beta A$. Also
$$
nA\left(1 + \mu A\right)^{n -1} = \alpha' + \beta' A
\quad\Longrightarrow\quad
nA\left(\alpha + \beta A\right) = \left(1 + \mu A\right)\left(\alpha' + \beta' A\right)
=
\alpha' + \beta' A + \mu\alpha' A
$$
$$
\mbox{That's means}\quad
0 = \alpha'\,,\quad n\alpha = \beta' + \mu\alpha'= \beta'\,,
\quad\mbox{with}\quad
\left.\alpha\right\vert_{\mu\ =\ 0} = 1\
\mbox{and}\ 
\left.\beta\right\vert_{\mu\ =\ 0} = 0\
$$
Then, $\alpha = 1$ and $\beta = n\mu$:
$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{\large\quad%
\left(1 + \mu A\right)^{n} \color{#000000}{\ =\ } 1 + n\mu A
\quad\color{#000000}{\Longrightarrow}\quad
\left(1 + A\right)^{n} \color{#000000}{\ =\ } 1 + nA
\quad}
\\ \\ \hline
\end{array}
$$
Or 'easy way':
$$
\left(1 + A\right)^{n}
=
1\ +\ \overbrace{\quad{n \choose 1}\quad}^{=\ n}\ A\
+\
\overbrace{\quad{n \choose 2}\,A^{2} + \cdots\quad}^{=\ 0}
$$
</s>