I am trying to prove that if $r$ is a nilpotent element, then $r+1$ is a unit, but I am having trouble factoring, which is extraordinarily embarrassing. If $r$ is nilpotent, then $r^n = 0$ for some $n$ and therefore $r^n+1=1$. I want to factor $r^n+1$ into $r+1$ times something else, but I am having trouble. From what I understand, $(r+1) \sum_{i=0}^{n-1} (-1)^{n-1-i} r^i$ should equal $1$, but
$$\begin{align} (r+1) \sum_{i=0}^{n-1} (-1)^{n-1-i} r^i & = \sum_{i=0}^{n-1} (-1)^{n-1-i} r^{i+1} + \sum_{i=0}^{n-1} (-1)^{n-1-i} r^i \\ & = \sum_{i=1}^{n} (-1)^{n-i} r^i + \sum_{i=0}^{n-1} (-1)^{n-1-i} r^i \\ & = r^n + \sum_{i=1}^{n-1} (-1)^{n-i} r^i +\sum_{i=1}^{n-1} (-1)^{n-1-i} r^i + (-1)^{n-1} \\ & = r^n + (-1)^{n-1}\end{align} $$
is clearly not $1$...What am I doing wrong!?! I realize this question has been asked before, but I am specifically concerned with the factorization part.
The answer should be reminiscent of the geometric series. Let $r$ be a nilpotent element. Then $\exists n \in \mathbb{N}$ such that $r^n = 0$. Then we also have that $(-r)^n = 0$, and hence, $$\begin{eqnarray} 1 &=& 1 - 0\\ &=& 1 - (-r)^n\\ &=& (1 + r)(1 - r + r^2 - \dots + (-1)^{n - 1} r^{n - 1}) \end{eqnarray}$$ And by closure, $(1 - r + r^2 - \dots + (-1)^{n - 1} r^{n - 1})$ is in the ring too so $(1 + r)$ is a unit.
EDIT: I just realized that I didn't address your own work. Notice that in your final step, $r^n + (-1)^{n - 1} = (-1)^{n - 1}$ since $r^n = 0$. If $n$ is even, then the inverse of $r + 1$ is simply $-\sum_{i = 1}^{n - 1} (-1)^{n - 1 - i}r^i$. If $n$ is odd, then we are done since the product is equal to 1!