Let $G$ be a cyclic finite group and $R$ a commutative unital ring. There is a morphism (I think it's called the augmentation morphism) $$\epsilon:R[G]\mapsto R\\ \sum\limits_{g\in G}c_gg\mapsto\sum\limits_{g\in G}c_g $$
The augmentation ideal $I(G)$ is defined as the kernel of $\epsilon$.
The problem is to that $I(G)$ is a principal ideal (it is generated by one element).
Since $G$ is cyclic and finite $\exists g\in G$ s.t. $G=\langle g\rangle=\{1_G,g,g^2,...,g^{|G|-1}$} and say $|G|=n$.
At first I tried to directly find the generating element by taking $I(G)\ni z=\sum\limits_{i=0}^{n-1}z_ig^i$ with the property that $h(z):=\max\{i:\ z_i\ne0\}$ is minimal. But then proving that every element of $I(G)$ is a multiple of $z$ is hard.
I also have a lead but don't know what to do with it:
If $x\in R$ and $\sum\limits_{g\in G}c_gg\in I(G)$ then $x=x-0=x-\sum\limits_{g\in G}c_g"="x-\sum\limits_{g\in G}c_g1_G$
As suggested by the comment of Mariano Suárez-Álvarez, the ideal is generated as an $R$-module by elements of the form $h - h'$. Since $G$ is cyclic we can write without loss of generality, for $m < n$, $$h - h' =g^m - g^n = \sum_{k=m}^{n-1} g^k - g^{k+1} = \sum_{k=m}^{n-1}g^k(1-g).$$ So the ideal is principal and generated by $1-g$.