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?
 A: Geometrically, your matrix represents a shear transformation that preserves the horizontal direction and shifts the vertical direction by the horizontal direction. What happens geometrically if you apply the same shear transformation $n$ times?
A: $$
\left(%
\begin{array}{cc}
1 & 1
\\
0 & 1
\end{array}\right)
=
1 + A
\quad\mbox{where}\quad
A
=
\left(%
\begin{array}{cc}
0 & 1
\\
0 & 0
\end{array}\right)
$$
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':
$$
\left(1 + A\right)^{n}
=
1\ +\ \overbrace{\quad{n \choose 1}\quad}^{=\ n}\ A\
+\
\overbrace{\quad{n \choose 2}\,A^{2} + \cdots\quad}^{=\ 0}
$$
A: 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.
A: 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
$$\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}^n=(I_2+N)^n=\sum_{k=0}^n {n\choose k}N^k={n\choose 0}I_2+{n\choose 1}N=I_2+nN=\begin{pmatrix} 1& n\\ 0 & 1 \end{pmatrix}$$
A: 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
$$ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^n = \exp \begin{pmatrix} 0 & n \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}. $$
A: 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!
A: Powers of matrices occur in solving recurrence relations.
If you write
$$
\begin{pmatrix} x_{n+1} \\ y_{n+1} \end{pmatrix} =
\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix} \begin{pmatrix} x_n \\ y_n \end{pmatrix}
$$
then clearly
$$
\begin{pmatrix} x_{n} \\ y_{n} \end{pmatrix} =
\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^n \begin{pmatrix} x_0 \\ y_0 \end{pmatrix}
$$
and also
$$
x_{n+1}=x_n+y_n, \qquad y_{n+1}=y_n
$$
from which you get
$$
x_n = x_0 + n\  y_0, \qquad y_n = y_0
$$
The first column of $A^n$ is given by taking $x_0=1$ and $y_0=0$, and so is $(1 \ 0)^T$.
The second column is given by taking $x_0=0$ and $y_0=1$, and so is $(n \ 1)^T$.
