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.

$(I+N)(I-N+N^2-N^3+...+(-1)^{k-1}N^{k-1})=I$
