Prove using induction $2^{3n}-1$ is divisble by $7$ for all $n$ $\in \mathbb N$ 
Show that 
$2^{3n}-1$ is divisble by $7$ for all $n$ $\in \mathbb N$

I'm not really sure how to get started on this problem, but here is what I have done so far:
Base case $n(1)$:
$\frac{2^{3(1)}-1}{7} = \frac{8-1}{7} = \frac{7}{7}$
But not sure where to go from here.  Tips?
 A: $$2^{3n}-1 = 8^n-1 = (8-1)(8^{n-1} + 8^{n-2} + \dots + 8^2 +8+1) = 7 \cdot (\mbox{something})$$
A: For simplicity, please note that $2^{3n}=8^n$.

First, show that this is true for $n=1$:


*

*$\frac{8^1-1}{7}=1\in\mathbb{N}$


Second, assume that this is true for $n$:


*

*$\frac{8^n-1}{7}=k\in\mathbb{N}$


Third, prove that this is true for $n+1$:


*

*$\frac{8^{n+1}-1}{7}=\frac{8\cdot8^n-1}{7}$

*$\frac{8\cdot8^n-1}{7}=\frac{8\cdot8^n-8+7}{7}$

*$\frac{8\cdot8^n-8+7}{7}=\frac{8(8^n-1)+7}{7}$

*$\frac{8(8^n-1)+7}{7}=\frac{8\cdot7k+7}{7}$ assumption used here

*$\frac{8\cdot7k+7}{7}=\frac{7(8k+1)}{7}$

*$\frac{7(8k+1)}{7}=8k+1\in\mathbb{N}$
A: Assume $2^{3n}-1$ is divisible by 7. Look at $2^{3(n+1)}-1$ and show that it is divisible by 7.
$2^{3(n+1)}-1 = 2^{3n+3}-1 = 8*2^{3n}-1 = (7+1)*2^{3n}-1 = (7*2^{3n} + 2^{3n})-1 = 7*2^{3n} + (2^{3n}-1)$ which is divisble by 7.
A: $2^{3n} = 8^{n}$. Let $s_{n} = 8^{n} - 1$.
$$8^{n + 1} - 1 = 8(8^{n}) - 1 = 8(8^{n} - 8) + 7 = 8(8^{n} - 1) + 7 = 8s_{n} + 7$$
