<s>[INST] Inverse of $I + A$ I am trying to solve the following exercise in Artin, without breaking into cases for even and odd $k$. 

A square matrix $A$ is called nilpotent if $A^k = 0$ for some $k > 0$. Prove that if $A$ is nilpotent, then $I + A$ is invertible. 

Here's my attempt. 
I claim that the inverse of $I + A$ is 
$$(I + A)^{-1} = \sum\limits_{n=0}^{k-1} (-1)^n A^n.$$
We prove this is a right inverse. 
\begin{align*}
(I + A)\sum\limits_{n=0}^{k-1} (-1)^n A^n & = \sum\limits_{n=0}^{k-1} (-1)^n (I + A)A^n \\
& = \sum\limits_{n=0}^{k-1} (-1)^n (A^n + A^{n+1}) 
\end{align*}
If $n$ is even, then $n + 1$ is odd and vice-versa. Hence, the series telescopes: 
\begin{align*}
& = (A^0 + A^{1}) - (A^1 + A^2) + (A^2 + A^3) + \ldots \pm (A^{k-1} + A^k) \\
& = I \pm A^k \\
& = I  
\end{align*}
It seems that the sign of the final term is dependent on whether $k$ is even or odd. However, it shouldn't matter, because if $A^k = 0$, then $-A^k = 0$. 
Is there a better way to formalize this fact? 
 [/INST] We know that, if $|x|<1$, then 
$$\dfrac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 -x^5 + \cdots$$
For just any square matrix, $\mathbf A$, it is not true that 
$$\dfrac{1}{1+\mathbf A} = 
  1 - \mathbf A + \mathbf A^2 - \mathbf A^3 + \mathbf A^4 -\mathbf A^5 + \cdots$$
But, since $\mathbf A$ is nilpotent, this becomes the finite sum
$$(1+\mathbf A)^{-1} = 
  1 - \mathbf A + \mathbf A^2 - \mathbf A^3 + \mathbf A^4 + \cdots  
  + (-1)^{k-1}\mathbf A^{k-1}$$
This can be easily verified by multiplying the right side by $1+\mathbf A$.
</s>