TITLE: Eisenbud 2.16 - units and nilpotents
QUESTION [5 upvotes]: This should be a pretty easy problem, but I'm a dummy so I'm stuck. Here's the statement:
Let $R$ be a $\mathbb Z$-graded ring, and $M$ a graded $R$-module, and let $x \in R_k$ for some non-zero integer $k$. Then $u = 1-x$ is not a zero divisor. Show that $u$ is a unit if and only if $x$ is nilpotent.
Now I know that a similar question has been asked here many times before, so let me say that I know how to show  $u$ is not a zero divisor, and I can show that if $x$ is nilpotent, $u$ is a unit. This is easy and has been done on this site a million times. My struggle is in the converse, that is to say, if $u$ is a unit, then I want to prove that $x$ is nilpotent. 
Apologies if this has also already been done on this site, but I can't seem to find the question on hand.

REPLY [4 votes]: Assume that $$(1-x)y=1$$ and let $y=\sum y_i$ be a sum of homogeneous elements $y_i$ then we have 
$$\sum y_i-xy_i=1$$
 Now we see that 
$$y_0=1$$ and that 
$$y_{i+k}=xy_i$$
Since the sum is finite, $x$ is nilpotent.
Note that if $-l$ is the smallest negative index where $y_{-l}$ is non zero then (assuming, $k$ positive by symmetry) we have
$$y_{-l}+xy_{-l}=0$$ and since these have different degrees 
$y_{-l}=0$.