Why is this the inverse of an nxn identity matrix plus an nxn upper triangular matrix? In class a few weeks we were told that about the inverse of $(I + N)$ is $(I - N + N^2 - \cdots)$ where $I$ is an $n\times n$ identity matrix and $N$ is an $n \times n$ upper triangular matrix with zeroes on the diagonal. We were told that this comes from elementary level math, "As we know from 5th grade..." so I don't feel comfortable asking the professor about it. Anyway where is this coming from? I do have a good deal of undergrad and grad level math (I just got into graduate school for math starting January) but I got a very sub-bar K-12 math education so some basics totally allude me to this day. 
 A: First note that $N$ is nilpotent, i.e., when computing the powers $N^2$, $N^3$, and so on, the upper triangle gets smaller and smaller until we find that $N^n$ is zero (and so are all higher powers). With this in mind,
$$ (I+N)(I-N+N^2-N^3\pm\cdots \pm N^n)=(I-N+N^2\mp\cdots +(-1)^n N^n)+N(I-N+N^2\mp\cdots +(-1)^n N^n)=(I-N+N^2\mp\cdots +(-1)^n N^n)+(N-N^2\pm\cdots +(-1)^n N^{n+1})=I$$
because all other summands cancel in pairs (apart from $N^{n+1}$ which is zero anyway)
A: "As we know from fifth grade" should not be taken literally. What is alluded to is a formula for the sum of an infinite geometric series:
$$
1 - r + r^2 -r^3 + r^4 - \cdots = \frac 1 {1+r}
$$
if $|r|<1.$
Maybe most people encounter that in high school.
The extent to which it can be applied to matrices is more complicated than $|r|<1.$ However, note that
\begin{align}
& (I+N) (I - N + N^2 - N^3 + N^4 - \cdots) \\[8pt]
= {} & I(I - N + N^2 - N^3 + N^4 - \cdots) \\
& {} +N(I - N + N^2 - N^3 + N^4 - \cdots) \\[8pt]
= {} & I - N + N^2 - N^3 + N^4 - \cdots \\
& \phantom{I}{} + N - N^2 + N^3 - N^4 + N^5 - \cdots
\end{align}
In the last row, one could say that each term cancels the one above it, so the sum is $I.$ That is problematic since questions of convergence are involved. In particular, what would happen with numbers with absolute values exceeding $1$ were there instead of matrices? To think about convergence one must first consider a finite sequence:
\begin{align}
& I - N + N^2 - N^3 + N^4 - \cdots \pm N^k \\
& \phantom{I} {} + N - N^2 + N^3 - N^4 + \cdots \mp N^k \pm N^{k+1} \\[12pt]
= {} & I \pm N^{k+1}
\end{align}
and then ask whether $N^{k+1}\to0$ as $k\to\infty.$
