Prove that if $n$ is a positive integer then $2^{3n}-1$ is divisible by $7$. I encountered a problem in a book that was designed for IMO trainees. The problem had something to do with divisibility. 

Prove that if $n$ is a positive integer then $2^{3n}-1$ is divisible by $7$.

Can somebody give me a hint on this problem. I know that it can be done via the principle of mathematical induction, but I am looking for some other way (that is if there is some other way)

$2^{3n} -1 = 8^n -1 = (7+1)^n -1 =$ (By Binomial theorem)$= C^n _0.7^n + C^{n}_{1} . 7^{n-1} + \dotsb + C^n_{n-1}7 + C^n_n -1 = 7.(C^n _0.7^{n-1} + C^{n}_{1} . 7^{n-2} + \dotsb + C^n_{n-1})$
