<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 statement is true for an arbitrary complex number $n$, with matrix logarithm. All the $\log$s below refer to the natural logarithm. 
For convenience, let's set $$ A = \begin{pmatrix} 0 & a \\ 0 & 0 \end{pmatrix}. $$
We know that if $k \in \mathbb  N$ and $k \ge 2$, then $A^k = O$.  Therefore,
$$ \exp A = \sum_{k=0}^\infty \frac{1}{k!}A^k = I + A. $$
All eigenvalues of $A$ are zero, so $A$ is the principal logarithm.
$$ \log \left( I + A \right) = A. $$
This leads to
$$ \log \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $$
And
$$ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^n = \exp \begin{pmatrix} 0 & n \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}. $$
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