For each $n\in \mathbf{N},2^{3n}-1$ is divisible by

 

We have 

$\begin{array}{l}{2}^{3n}-1={\left(2^3\right)}^n-1=(1+7{)}^n-1\\ =1{+}^n{C}_1\left(7\right){+}^n{C}_2\left(7^2\right)+\dots {+}^n{C}_n\left(7^n\right)-1\\ =7m\end{array}$

Where $m{=}^n{C}_1{+}^n{C}_2\left(7\right)+\dots {+}^n{C}_n\left(7^{n-1}\right)$

is a positive integer.

Thus,  $7$ divided $2^{3n}-1$ for all $n\in \mathbf{N}$