Given a matrix $A$ find $A^n$. $A=$$
    \begin{bmatrix}
    1 & 2\\ 
    0 & 1  
    \end{bmatrix} $
Find $A^n$. 
My input: 
$A^2= \begin{bmatrix}
    1 & 2\\ 
    0 & 1  
    \end{bmatrix} 
\begin{bmatrix}
    1 & 2\\ 
    0 & 1  
    \end{bmatrix}  = \begin{bmatrix}
    1 & 4\\ 
    0 & 1  
    \end{bmatrix} $
$A^3 = \begin{bmatrix}
    1 & 6\\ 
    0 & 1  
    \end{bmatrix} $
......
$A^n = \begin{bmatrix}
    1 & 2n\\ 
    0 & 1  
    \end{bmatrix} $
This was very basic approach. I want to know if there is any other way a smart trick or something to solve this problem ?  

You can use this too
$$A=\begin{pmatrix}1&2\\0&1\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}+\begin{pmatrix}0&2\\0&0\end{pmatrix}$$
$$A=I_2+B$$
And B is a nilpotent matrix $\implies B^2=0$
$$A^n=(I_2+B)^n$$
Use  binomial theorem
