An identity in Ring of characteristic $p$ prime Is it true that in a ring of prime characteristic $p$ results that
$(x-1)^{p-1}=1+x+x^2+...+x^{p-1}$ ?
If this is not true in general, the assumption that $x$ is a nilpotent element (let's say $x^{p^n}=0$) make it works?

Yes this is true, the binomial theorem gives
$$(x - 1)^{p-1} = \sum_{k = 0}^{p-1} \binom{p - 1}{k}(-1)^{p - 1 - k}x^k$$
and in characteristic $p$ one can show that
$$\binom{p - 1}{k} = (-1)^k$$
and
$$(-1)^{p - 1 - k} = (-1)^k$$
so the above simplifies to
$$(x - 1)^{p-1} = \sum_{k = 0}^{p-1}x^k.$$
