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 ?  

What you did was the smart approach. That is, you computed the first few terms of the sequence $(A^n)_{n\in\mathbb N}$ and you noticed a patern. The only thing that remains to be done is to prove that the pattern is real, but that's easy. Obviously,$$A^1=A=\begin{pmatrix}1&2\\0&1\end{pmatrix}$$and$$A^n=\begin{pmatrix}1&2n\\0&1\end{pmatrix}\implies A^{n+1}=A.\begin{pmatrix}1&2n\\0&1\end{pmatrix}=\begin{pmatrix}1&2(n+1)\\0&1\end{pmatrix}.$$
