Why does the congruence hold? Let $p$ be a prime number and $\mathbb{Q}(\zeta)$ be the pth cyclotomic number field where $\zeta$ is any primitive pth root of unity.
Writing $$t=b_0+b_1\zeta+...+b_{p-2}\zeta^{p-2} $$ with $b_j \in \mathbb{Z}$ , we get $$t^p \equiv b_0^p+b_1^p+...+b_{p-2}^p \pmod{p\mathbb{Z}[\zeta]}$$
I understand the congruence if I consider this modulo p .

First prove that $t^p = b_0^p + (b_1\zeta)^p + \cdots + (b_{p-2}\zeta^{p-2})^p$.
Hint: write it out and note what terms get a coefficient divisible by $p$, then note that all those coefficients are in your ideal $(p)$.
Another hint: Maybe start small and show $(a + b)^p = a^p + p(\cdots) + b^p$.
Next, use $\zeta^p = 1$ to arrive at your final answer.
