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

Question

How to prove that: $$ 8^{18}-1\equiv0\pmod7 $$ In the simplest way?

Answer 1

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

Answer 2

Hint: $$ 8 \equiv 1 \mod 7\ \ \text{and}\ \ a^n \equiv b^n \mod m$$ Given any $a,b \in \mathbb{Z}, \ \ m > 1$ and $n \geq 1$ such that $a \equiv b \mod m$.

Answer 3

Ok, I found overall formula: $$ a^n \equiv b^n \pmod {a-b} $$ hence: $$ 8^{18} \equiv 1^{18} \pmod {8-1} $$

$$ 8^{18} - 1 \equiv 0 \pmod {8-1} $$

Answer 4

Alternatively, you can just note that $8^{18}-1=7\cdot2573485501354569$.