<s>[INST] 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?
 [/INST] 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

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

*$\frac{7(8k+1)}{7}=8k+1\in\mathbb{N}$
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