Matrices in Geometric progression 
Let p be a non singulqar matrix, and $\mathrm{I + p+ p^2 +...+p^n = O}$, find $p^{-1}$

Attempt: 
$\mathrm{p +p^2 +...+p^n = -I \implies p^{-1}= -(I+p+ p^2 +...+ p^{n-1})}$
I am not sure how to continue from there because how can the GP formula be valid for matrices becuase their division is not defined. 
 A: With
$p^n + p^{n - 1} + p^{n -2} + \ldots + p^2 + p + I = 0 \tag 1$
we may write
$p(p^{n - 1} + p^{n -2} + \ldots + p^2 + p + I) = -I, \tag 2$
which shows that
$p^{-1} = -(p^{n - 1} + p^{n -2} + \ldots + p^2 + p + I). \tag 3$
Note the highest degree in $p$ occurring on the left is $n - 1$; the "attempt" in the original text of the question erred in allowing a term of degree $n$ to appear in the given polynomial expression for $p^{-1}$; but I see this has now been corrected.  
If one seeks to follow through with the analogy to geometric series of numbers, and write
$p^{n - 1} + p^{n -2} + \ldots + p^2 + p + I = (p^n - I)(p - I)^{-1}, \tag 4$
one has to take care that $p - I$ is in fact itself invertible; given that this is the case, then (4) follows directly from
$(p - I)(p^{n - 1} + p^{n -2} + \ldots + p^2 + p + I)  = p^n - I, \tag 5$
which is well-known, and in any event may be proved by a simple inductive argument which I leave to the reader along with the invitation to shoot me a comment if there are any further questions.
In fact, we may also write
$p^{-1} = p^n, \tag 6$
since from (1), 
$p^n = -(p^{n - 1} + p^{n - 2} + \ldots + p + I); \tag 7$
then combining this with (3) we arrive at (6). 
A: A slightly different argument -
We have $p^n+p^{n-1}+\ldots+I=0$. Since we know that $p\neq I$ (Why?), we multiply the equation by $p-I$, which gives
$p^{n+1}-I=0\Rightarrow p^{-1}=p^n$
