Prove using mathematical induction that $2^{3n}-1$ is divisible by $7$ So, i wanna prove $2^{3n}-1$ is divisible by $7$, so i made this:
$2^{3n}-1 = 7\cdot k$ -> for some $k$ value
$2^{3n+1} = 1+2\cdot1 - 2\cdot1 $
$2^{3n+1} - 1-2\cdot1 + 2\cdot1 $
$2^{3n}\cdot2 - 1-2\cdot1 + 2\cdot1$
$2(2^{3n}-1) -1 +2$
$2\cdot7k+1$ -> made this using the hypothesis.
so, i dont know if its right, or if its wrong, i dont know how to keep going from this, or if its the end.
Thanks.

(Without induction)
There is a very useful identity $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \dots + a b^{n-2} + b^{n-1})$.
If you take $a = 2^{3} = 8$ and $b=1$, the result becomes obvious.
