Prove that if $A$ is an $n \times n$ matrix such that $A^{4} = 0$ then $(I_n - A)^{-1}=I_n+A+A^2+A^3$ Prove that if $A$ is an $n \times n$ matrix such that $A^{4}$ = 0 then:
$$(I_n - A)^{-1}=I_n+A+A^2+A^3$$
My proof is as follows:
$$(I_n - A)(I_n - A)^{-1}=I_n$$
$$(I_n - A)^{-1}=I_n/(I_n - A)$$
$$I_n/(I_n - A)=I_n+A+A^2+A^3$$
$$I_n=(I_n - A)(I_n+A+A^2+A^3)$$
$$I_n=I_n+A+A^2+A^3-A-A^2-A^3-A^4$$
$$I_n=I_n-A^4$$
because we know that: $$A^4=0$$
therefore:
$$I_n=I_n$$
Is this an acceptable justification or have I made an error in my logic?
*I apologize for any poor formatting

Hint: Let $B=I+A+A^2+A^3$. Compute $AB$.
