Answer
${A^n} = \left[ {\begin{array}{*{20}{c}}
1&n\\
0&1
\end{array}} \right]$
Work Step by Step
Let us first determine A^2
${A^2} = AA = \left[ {\begin{array}{*{20}{c}}
1&1\\
0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&1\\
0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{1(1) + 1(0)}&{1(1) + 1(1)}\\
{0(1) + 1(0)}&{0(1) + 1(1)}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&2\\
0&1
\end{array}} \right]$
Next determine A^3
${A^3} = {A^2}A = \left[ {\begin{array}{*{20}{c}}
1&2\\
0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&1\\
0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{1(1) + 2(0)}&{1(1) + 2(1)}\\
{0(1) + 1(0)}&{0(1) + 1(1)}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&3\\
0&1
\end{array}} \right]$
Next determine A^4
${A^4} = {A^3}A = \left[ {\begin{array}{*{20}{c}}
1&3\\
0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&1\\
0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{1(1) + 3(0)}&{1(1) + 3(1)}\\
{0(1) + 1(0)}&{0(1) + 1(1)}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&4\\
0&1
\end{array}} \right]$
We note that the matrix A^n has 1 on the diagonal, 0 in the bottom left corner and n in the upper right corner.
So,
${A^n} = \left[ {\begin{array}{*{20}{c}}
1&n\\
0&1
\end{array}} \right]$
