So now working up to nilpotent groups, there is this problem in my book where its given as a problems to think about , and it goes:

$R$ is a commutative ring with unity $1$. Prove that if $J$ is defined by $K=\{{1-s|s\in R

\text{ nilpotent}\}}$, then it is a group under multiplication. Now suppose $R$ is finite with $|R|=n$, and $s$ nilpotent, show that $(1-s)^n=1$.

How I tried: So for the group part I think the inverse is pretty easy to find which is just $(1-s)(1 + s + s^2+ \cdots +s^{n-1}) = 1-s^n = 1 \rightarrow$ Inverse = $1+s+s^2+\cdots+s^{n-1}$. Im stuck at trying to show that there exists some 1-t, t\in R, then $(1-t)(1-s)=1-((t+s)-st)$, how do I show that $(t+s)-st$ is nilpotent?

For the second problem I think I can see that $|K| | |R|$ should be true, but I dont know why this is the case, but clearly $K$ is a coset of the nilpotent $s$ so is this why the order of $K$ divides $R$? But why is $|K|$ diving $|R$| show that $(1-s)^n =1$ ? Could anyone show me how to write this formally so it makes sense in a more coherent manner?