On p. 100 of "Classical Theory of Algebraic Numbers" from Ribenboim there is a proof that $\{1,...,\zeta^{p-2}\}$ is an integral basis of $\mathcal{O}_K$, where $K=\mathbb{Q}$, $\mathcal{O}_K$ being the ring of algebraic integers and $\zeta$ a primitive pth root of unity.
In one line we have $x\zeta^{p-j}=a_0\zeta^{p-j}+a_1\zeta^{p-j+1}+...+a_{j-1}\zeta^{p-1}+a_j+a_{j+1}\zeta^{p-1}+...+a_{p-2}\zeta^{p-j-2}$, with $x\in\mathcal{O}_K$, $a_i\in\mathbb{Q}$. Then he writes that by expressing $\zeta^{p-1}$ in terms of lower powers of $\zeta$ we may write $x\zeta^{p-j}=(a_j-a_{j-1})+a'_1\zeta+...+a'_{p-2}\zeta^{p-2}$.
But I cannot see how we get this or what he exactly means by expressing $\zeta^{p-1}$ in terms of lower powers of $\zeta$. So an explanation for this step would be very helpful.
Since $\;\zeta\;$ is a complex non-real root of $\;x^p-1=(x-1)(x^{p-1}+x^{p-2}+\ldots+x+1)\;$ , it then is a root of the second factor, so
$$\zeta^{p-1}+\zeta^{p-2}+\ldots+\zeta+1=0\implies \zeta^{p-1}=-\sum_{k=1}^{p-2}\zeta^k$$