modular arithmetic - How to prove that $8^{18} - 1$ is divisible by $7$

How to prove that:

$$8^{18}-1\equiv0\pmod7$$
In the simplest way?

Answer

Yet another one: $a^{18}-b^{18}=(a-b)\left(a^{17}+a^{16}b+\cdots+ab^{16}+b^{17}\right) \ ,$ hence

$$8^{18}-1=(8-1)\left(8^{17}+8^{16}+\cdots+8+1\right),\quad\text{which is a multiple of $7$.}$$