Suppose $A^n = 0$ matrix for some $n > 1$. Find an inverse for $I - A$. Source: Linear Algebra by Lay (4 edn 2011). p. 160. Chapter 2 Supplementary Exercise 4.

Solution: From p. 160 Supplementary Exercise 3, the inverse of $I-A$ is probably $I+A+A^{2}+...+A^{n-1}$. To verify this, compute
  $ (I \color{orangered}{-A} )\color{forestgreen}{(I+A+\cdots+A^{n-1})}=I+A+\cdots+A^{n-1} \quad \color{orangered}{-A}(I+A+\cdots+A^{n-1})=I \color{orangered}{-A} A^{n-1}=I-A^{n}. \;[...] \; \blacksquare$

How can you divine that the inverse of $I-A$ is  $\color{forestgreen}{\sum_{0 \le i \le n - 1} A^{i}}$ ? The solution doesn't explain.
I already understand that $ (I \color{orangered}{-A} )\color{forestgreen}{(I+A+\cdots+A^{n-1})}= ... =I-A^{n}. $ 
 A: Think of the geometric series. For $|q|<1$: $\frac {1}{1-q}=1+q+q^2+\cdots $. Now replace $q$ by $A$, think of $1$ as an identity thingy $I$,  and use $^{-1}$ for the inverse: $(I-A)=I+A+A^2+A^3\cdots$. Now, forget about the fact that $A$ is supposed to be a number and that the equality only holds if $|A|<1$, and just concentrate on the algebra for matrices. There is an immediate problem here namely that the RHS is an infinite sum so it makes no sense without some limiting process. But, what if $A^n=0$? Well, then the sum on the right is finite, and there is no problem. 
This is a way to motivate the inverse for $I-A$ as that sum. It is far less silly than it might initially appear since in fact, an almost trivial proof shows that if the infinite series $I+A+A^2+\cdots +A^n +\cdots $ converges (in whatever sense, but typically in the operator norm, i.e., strong convergence), then its sum is indeed the inverse of $I-A$. This series is called the Neumann series of $A$ and is a very important tool in, for instance, solving differential equations. Incidentally, this is most useful in infinite dimensional vectors spaces where $A$ is a linear operator, not a matrix. Then you do have to check that the inverse is two sided since for infinite dimensional vector spaces the existence of a one-sided inverse does not imply invertibility. 
