Answer     8

Because $\varphi(7)=6$ and $(10, 7)=1$, we find $10^6\equiv 1\pmod{7}$ by Euler's theorem.

Meanwhile by the conclusion of # 4224, we have $10^k\equiv 4\pmod{6}$ hold for any integer $k$. Therefore $$10^{10^m}\equiv 10^{6n} \times 10^4\equiv 10^4\pmod{7}$$

holds for any positive integer $m$ and corresponding positive integer $n$. It follows that $$\begin{array}{ll} & 10^{10}+10^{100}+10^{1000}+\cdots+10^{\overbrace{\scriptsize{10\cdots 0}}^{2018}} \\ \equiv & 10^4 + 10^4 + \cdots + 10^4 \\ \equiv & 10^4\times 2018 \\ \equiv & 3^4 \times 2 \\ \equiv & 4\times 2 \\ \equiv & \boxed{1}\pmod{7}  \end{array}$$