<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] Use induction on $n$. 
(1) Prove the base case (trivial), perhaps even establish the case for $n = 2$ (two base cases here are not necessary, but as you found, it helps reveal the pattern.) 
(2) Then assume it holds for $n = k$. 
(3) Finally, show that from this assumption, it holds for $n = k+1$.

You've established the base case(s). Now, (2) assume the inductive hypothesis (IH) $$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^k = \begin{pmatrix} 1 & k \\ 0 & 1\end{pmatrix}.$$
Then, $$\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}^{k + 1} = \begin{pmatrix} 1 & 1\\ 0 & 1\end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^k \quad \overset{IH}{=} \quad \begin{pmatrix} 1 & 1\\ 0 & 1\end{pmatrix} \begin{pmatrix} 1 & k \\ 0 & 1\end{pmatrix}=\quad\cdots$$
I think you can take it from here!
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