Let us conventionally set $x_{2018}=x_{1}$. For all $i=1, \ldots, 2017$, we have $x_{i+1}-$ $x_{i}=x_{i}\left(x_{i}-1\right)+1-x_{i}=\left(x_{i}-1\right)^{2} \geqslant 0$, so $x_{1}=x_{2018} \geqslant x_{2017} \geqslant \cdots \geqslant x_{1}$. We deduce that the $x_{i}$ are all equal, hence $0=x_{i+1}-x_{i}=\left(x_{i}-1\right)^{2}=0$. Therefore, $x_{i}=1$ for all $i$. Conversely, it is clear that $x_{i}=1$ is a solution.