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?
