<s>[INST] Proving that $ (A + I)^n=(2^n-1)A + I $ Assume that  $ A^2 = A$ , where $A$ is  a square matrix. 

Then prove $(A + I)^n=(2^n-1)A + I $

Could someone help ?
 [/INST] By indution if $(A+I)^n=(2^n-1)A+I$ and $A^2=A$, we have
\begin{eqnarray}
(A+I)^{n+1}&=&(A+I)^n(A+I)=((2^n-1)A+I)(A+I)\\
&=&(2^n-1)A^2+A+(2^n-1)A+I\\
&=&2(2^n-1)A+A+I\\
&=&(2^{n+1}-2)A+A+I\\
&=&(2^{n+1}-1)A+I
\end{eqnarray}
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