One of $2^1-1,2^2-1,...,2^n-1$ is divisible by $n$ for odd $n$ 
Let $n$ be an odd integer greater than 1. Show that one of the numbers $2^1-1,2^2-1,...,2^n-1$ is divisible by $n$.

I know that pigeonhole principle would be helpful, but how should I apply it? Thanks.

Hint: We know that $\gcd(2,n)=1$
$2^{\phi(n)} \equiv 1\pmod n$
And $1<\phi(n)<n$ and we already have $2^1-1 \equiv 1 \pmod n$
You can use pigeon-hole principle after this.
Another Approach: 
Actually, the $\phi(n)$ or $\lambda(n)$ occurs just between $1$ and $n$, therefore,
$2^{\lambda(n)} \equiv 1 \pmod n  \implies 2^{\lambda(n)} -1 \equiv 0 \pmod n$.
This involves no Pigeons or Pigeon-holes.:)
Here $\lambda(n)$ is smallest number which occurs before $\phi(n)$ such that $a^{\lambda(n)} \equiv 1 \pmod n$. Also known as Carmichael Number.
