<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] The matrix
$$N=\begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}$$
is nilpotent with index 2 of nilpotency: $N^2=0$ so by the binomial formula we have
$$\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}^n=(I_2+N)^n=\sum_{k=0}^n {n\choose k}N^k={n\choose 0}I_2+{n\choose 1}N=I_2+nN=\begin{pmatrix} 1& n\\ 0 & 1 \end{pmatrix}$$
</s>