Finding the inverse of $I+N$ where $N$ is nilpotent Suppose that $N \in M_{n,n}(\mathbb{C})$ is nilpotent (that is, $N^k = 0$ for some integer $k > 0$). Show that $I+N$ is invertible, and find its inverse as a polynomial in $N$.
I think I got the first part down "intuitively". Noticing that $N$ is nilpotent, so $N$ will be a matrix with a diagonal (any diagonal) with just $1$'s as its entries. Then if I add the identity matrix to it, the diagonal will definitely have $1$ as a diagonal, and so $\det(N+I) = 1$ hence invertible. Is there a more standard way to prove this rather than just "talking" through it?
Also, I'm unsure how I could approach the second part.
 A: $(I+N)(I-N+N^2-N^3+...+(-1)^{k-1}N^{k-1})=I$
A: Something that comes up over and over again in different parts of math is the geometric series.  For motivation, consider the infinite sum
$$S = 1 + x + x^2 + \cdots$$
Here $x$ could be a complex number, a matrix, a linear operator, or something else.  For the moment, just pretend the infinite sum makes sense.  To find a closed expression for $S$, notice that $S$ appears on the right hand side of the equation again if you factor out $x$:
$$1 + x + x^2 + \cdots = 1 + x(1+ x + \cdots) = 1 + xS$$
Now you can solve for $S$ as 
$$S = \frac{1}{1-x}$$
In particular, $S$ is the inverse of $1-x$.  As I mentioned, the infinite sum $S$ may or may not make sense.  If $x$ is an element of a metric space in which addition is defined, the usual definition of $S$ is the limit as $n$ goes to infinity of $1 + x + x^2 + \cdots + x^n$, provided this limit exists.  If $x$ is a nilpotent matrix, then $S$ also makes sense, because it is just a finite sum.  Reasoning as I've done above will produce for you an inverse of $1+x$.
