Let $p$ be the smallest prime divisor of $n$ and let $k$ be the order of $n$ mod $p$. Then of course: $k>1$, $k|n$, $k|p-1$ which implies that $n$ has a prime divisor smaller than $p$, contradiction.

Solution by Christian Hirsch (similar, but still different): Assume that $n$ is the smallest number $>1$ with that property. Then let $k = \gcd(n,\varphi(n)) <n$ and see that $2^k \equiv 1 \mod n$ (by $2^n , 2^{\varphi(n)} \equiv 1 \mod n$), thus $k|2^k-1$. By minimality we have $k=1$, giving $2 \equiv 1 \mod n$, contradiction.

We apply Fermat’s method of infinite descent to the prime factors of n . Let p1 be prime divisor of n and q be the smallest positive integer for which p1 divides 2q-1. From Fermat’s Little theorem it follows that p1 divides 2p1-1-1. Also we have q< p1-1<p Let us prove that q divides n. If not let n=kq+r, where 0<r<q. Then 2n-1=2kq2r-1=(2q-1+1)2r-1 which is obviously equal to 2r-1 modulo p1. It follows that p1 divides 2r-1, contradicting the minimality of q. Hence q divides n and 2q<p1. Let p2 be a prime divisor of q. Then it is also a divisor of n and hence p2<p1. Repeating the argument we construct an infinite sequence of prime divisors. Hence the proof follows. This problem is from the 1st Putnam in 1939 most probably!

Let $ p$ be the smallest prime divisor of $ n$. So we have $ 2^n\equiv_p1$, which means that $ ord_p2|n$, but by the Fermat's Little Theorem, $ \mbox{ord}_p2\le p - 1$, so $ \mbox{ord}_p2 = 1$, so $ 2\equiv_p1$, which is impossible.

Peter wrote: Let $n$ be an integer with $n \ge 2$. Show that $n$ does not divide $2^{n}-1$. Assume that $n|2^n-1$. Let $p$ be the smallest prime factor of $n$. Therefore $2^n-1 \equiv 0 \pmod{p}$. By Fermat's Little Theorem we have $2^{p-1} \equiv 1 \pmod{p}$. Thus, $\text{ord}_p(2)|n$ and $\text{ord}_p(2)|p-1$. It follows that $p>\text{ord}_p(2)$. If $\gcd ( \text{ord}_p(2),n)>1$ then $r|n, r| \text{ord}_p(2)$. It follows that $r|n$ and $p>\text{ord}_p(2)>r$, a contradiction since $p$ is the smallest prime factor of $n$. Thus $\gcd (n, \text{ord}_p(2))=1$. But $\text{ord}_p(2)|n$ then $\text{ord}_p(2)=1$. Hence $p|2-1$ or $p|1$, a contradiction. Thus, $n \nmid 2^n-1$ with $n \ge 2$.

It's from Putnam.

I apologize to everyone for my solution being very similar to that of ZetaX. We suppose that there is $n\ge 2$ s.t. $n | 2^n-1$.We consider the minimum such $n$.Obviously $n$ is odd.Then by Euler's theorem,$2^{\varphi (n)}\equiv 1$ mod $n$. Let the order of $2$ in mod $n$ be $k$.$k$ is common divisor of $n$ and $\varphi (n)$.Thus $2^k\equiv 1$ mod $k$.Since $k\le \varphi (n)<n$ and minimality of $n$,then $k=1$.Hence $2\equiv 1$ mod $n$ which is absurd.Therefore $n\nmid 2^n-1(n\ge 2)$.$\blacksquare$

Peter wrote: Let $n$ be an integer with $n \ge 2$. Show that $n$ does not divide $2^{n}-1$. Assume that,p is the least prime factor of $2^n-1$.p here can't be equal to 2 since $2^n-1$ is a odd number.so,$p>2$ by FLT,$2^{p-1} \equiv 1(modp)$ By fundamental theorem of order,we can say,$Ord_{p}{2}\mid n$ On the other hand,$Ord_{p}{2}\leq {p-1}$ which contradicts our assumption that,p is the least prime factor of n. So,there does'nt exist such n. Now we are done.

This is clearly true for even $n$. Now take odd $n$ and suppose that $n\mid 2^n-1$. Suppose $p>2$ is the smallest prime divisor of $n$. Then $$2^n\equiv 1\pmod p \implies \text{ord}_p(2)\mid n,$$a contradiction since $1<\text{ord}_2(p)<p$.