Proof of n-times matrix multiplication I want to proof the following: $$
  \left[ {\begin{array}{cc}
    1 & 0 \\
    1 & 2 \\
  \end{array} } \right]^n=  \left[ {\begin{array}{cc}
    1 & 0 \\
    2^n-1 & 2^n \\
  \end{array} } \right]
$$
I betitle the matrix on the left with $A$ and the matrix on the right with $B$.
I am not finding the right idea. I started with stating that the resulting matrix of $
  \left[ {\begin{array}{cc}
    1 & 0 \\
    1 & 2 \\
  \end{array} } \right]^n
$ has to be a $2\times2$ matrix because of the definition of matrix multiplication. Then I wrote the following
$$  \left[ {\begin{array}{cc}
    1 & 0 \\
    1 & 2 \\
  \end{array} } \right]^n=  \left[ {\begin{array}{cc}
    1 & 0 \\
    1 & 2 \\
  \end{array} } \right]\times  \left[ {\begin{array}{cc}
    1 & 0 \\
    1 & 2 \\
  \end{array} } \right]\times\dots\times  \left[ {\begin{array}{cc}
    1 & 0 \\
    1 & 2 \\
  \end{array} } \right]=  \left[ {\begin{array}{cc}
    1\times 1+0\times 1 & 1\times 0 + 0\times 2 \\
    1\times 1 + 2\times 1 & 1\times 0 + 2\times 2 \\
  \end{array} } \right]\times\dots\times  \left[ {\begin{array}{cc}
    1 & 0 \\
    1 & 2 \\
  \end{array} } \right]=  \left[ {\begin{array}{cc}
    1 & 0 \\
    3 & 4 \\
  \end{array} } \right]\times\dots\times  \left[ {\begin{array}{cc}
    1 & 0 \\
    1 & 2 \\
  \end{array} } \right]$$
My idea behind this was that this should show that the first line will never change, because we always just multiply with the same matrix again. I think this is quite easy to see.
For the second line it's in my opinion quite easy to see that the second number ($b_{22}$) is always the old second number ($a_{22}$) times 2 and because the first number with wich we started ($a_{22}$) is the number 2 it's just always powers of 2. And the first number ($b_{21}$) is always the old first number ($a_{21}$) plus the old second number ($a_{22}$). And because the first number with wich we started ($a_{21}$) is on less than the second number with wich we started ($a_{22}$) it's always one less than the power of 2.
I have problems with explaining my proof and I am not quite sure if it's correct, so can maybe someone recommend a more exact way of proofing this, because I don't know if my sentences are exact enough.
Induction Try
Thanks to the comment from @Jaap Sherphuis I now tried to proof the statement with induction and started with the basecase $n=1$.
$$  \left[ {\begin{array}{cc}
    1 & 0 \\
    1 & 2 \\
  \end{array} } \right]^1=\left[ {\begin{array}{cc}
    1 & 0 \\
    2^1-1 & 2^1 \\
  \end{array} } \right]$$
This is obviously true. Now I take $
  \left[ {\begin{array}{cc}
    1 & 0 \\
    1 & 2 \\
  \end{array} } \right]^n=  \left[ {\begin{array}{cc}
    1 & 0 \\
    2^n-1 & 2^n \\
  \end{array} } \right]
$ as given and try to show with it the statement for $n=n+1$.
$$
  \left[ {\begin{array}{cc}
    1 & 0 \\
    1 & 2 \\
  \end{array} } \right]^{n+1}=  \left[ {\begin{array}{cc}
    1 & 0 \\
    2^{n+1}-1 & 2^{n+1} \\
  \end{array} } \right]\Leftrightarrow
  \left[ {\begin{array}{cc}
    1 & 0 \\
    1 & 2 \\
  \end{array} } \right]^n \times   \left[ {\begin{array}{cc}
    1 & 0 \\
    1 & 2 \\
  \end{array} } \right]=  \left[ {\begin{array}{cc}
    1 & 0 \\
    2^n-1 & 2^n \\
  \end{array} } \right] \times   \left[ {\begin{array}{cc}
    1 & 0 \\
    1 & 2 \\
  \end{array} } \right]
$$
Because $
  \left[ {\begin{array}{cc}
    1 & 0 \\
    1 & 2 \\
  \end{array} } \right]^n=  \left[ {\begin{array}{cc}
    1 & 0 \\
    2^n-1 & 2^n \\
  \end{array} } \right]
$ is true the whole statement is true for $n=n+1$ and thus the whole statement is true for every $n\in\mathbb{N}$.
Is this the correct approach?

The matrix can be represented as $$I+P,\quad P=\begin{pmatrix}0&0\\ 1&1\end{pmatrix}$$ We have $P^2=P.$ Thus $$(I+P)^n=I+\sum_{k=1}^n{n\choose k}P^k=I+(2^n-1)P$$
