2018 again? #2

$2018 \equiv 2 \text{ mod 8} \Rightarrow 2018^n \equiv 2^n \text{ mod 8}$ , but if $2018^n$ ends at $2018$ then $2018^n \equiv 2 \text{ mod 8}$ beacuse all the numbers ending at 2018 are congruent with 2 mod 8, and hence, $2^n \equiv 2 \text{ mod 8}$ , and this implies $n = 1$ , so there doesn't exist $n > 1$ such that $2018^n$ ends at 2018.

Even, there doesn't exist $n > 1$ such that $2018^n$ ends at 18.