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)
Hint: Note that $8 \equiv 1~~~(\text{mod } 7)$. 
So,
$$2^{3n}=(2^3)^n=8^n\equiv \ldots~~~(\text{mod } 7)=\ldots~~~(\text{mod } 7)$$
Try to fill in the gaps!

Solution:
Note that $8 \equiv 1~~(\text{mod } 7)$. This means that $8$ leaves a remainder of $1$ when divided by $7$. Now assuming that you are aware of some basic modular arithmetic,
$$2^{3n}=(2^3)^n=8^n\equiv 1^n ~~(\text{mod } 7)=1~~(\text{mod } 7)$$
Now if $2^{3n}\equiv 1~~(\text{mod } 7)$ then it follows that,

$$2^{3n}-1=8^n-1\equiv (1-1)~~(\text{mod } 7)~\equiv 0~~(\text{mod } 7)\\ \implies 2^{3n}-1\equiv 0~~(\text{mod } 7)$$
Or in other words, $2^{3n}-1$ leaves no remainder when divided by $7$ (i.e. $2^{3n}-1$ is divisible by $7$). As desired
Use Proof by induction.It's easy to show that if this holds for "n" ,then it holds for "n+1" and of course it's true for n=1