Proving that $I+A$ is invertable when $A$ is nilpotent: What intuition leads to a particular approach? In an answer to this question, it has been suggested to consider the following: $$(I+A)(\sum_{j=0}^n(-A)^j)$$
Through a series of algebraic operations, it can be shown that $\sum_{j=0}^n(-A)^j$ is in fact the inverse of $I+A$.
How would we have known to multiply by $\sum_{j=0}^n(-A)^j$? If there isn't an identity or formula that would indicate such a multiplication is a reasonable avenue of inquiry, then how would we otherwise derive $\sum_{j=0}^n(-A)^j$?
 A: If $z\in\mathbb C$ and $\lvert z\rvert<1$, then$$\frac1{1+z}=1-z+z^2-z^3+\cdots$$In other words,$$(1+z)\left(1-z+z^2-z^3+\cdots\right)=1.$$This, together with the fact that $A^n=0$ if $n\gg1$, should make you think that it would be a good idea to try to prove that an inverse of $\operatorname{Id}+A$ is $1-A+A^2-A^3+\cdots$ (which is a finite sum, in this case).
A: I guess a natural way of thinking about this stems from the formula for a geometric series for real numbers: if $|x| < 1$, then
\begin{align}
\sum_{j=0}^\infty x^j = \dfrac{1}{1-x}
\end{align}
Now, replacing $x$ with $-x$ gives the formula
\begin{align}
(1+x)^{-1} = \dfrac{1}{1+x} = \sum_{j=0}^{\infty} (-x)^j
\end{align}
So, it might be natural to try this for matrices as well. And now if $A$ is nilpotent, then on the RHS, we only have a finite number of summands (in fact at most $n$ summands if $A$ is $n \times n$). Hence, one might expect that $\sum_{j=0}^n (-A)^j$ is the inverse matrix of $I+A$. Then, a simple computation verifies that it actually is.
A: Recall that 
$$x^{2n+1}+y^{2n+1}=(x+y)(x^{2n}-x^{2n-1}y+x^2y^{2n-2}-..+y^{2n})$$
Now, since $I$ and $A$ commute, the same formula holds for $x=I$ and $y=A$. Therefore, 
$$I+A^{2n+1}=(I+A)(I-A+A^2-A^{3}+...+A^{2n})$$
Now use the fact that $A^n=0$ implies 
$$A^{n}=A^{n+1}=...=A^{2n+1}=0$$
