For $n \geq 2$, show that $n \nmid 2^{n}-1$ Here is a problem which i have not been able to do for quite sometime.


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*For $n \geq 2$, show that $n \nmid 2^{n}-1$.


I have thought of proving this in two ways: One by using induction which didn't actually work. Next by Fermat's little theorem we have $2^{p-1} \equiv 1 \ (\text{mod} \ p)$, which actually says that for $p \mid 2^{p}-2$. But i couldn't proceed more than this. Any ideas by which i can actually solve the problem?
 A: Hint $  $ prime $\, p\mid n\mid 2^{n}\!-1\ \Rightarrow\bmod p\!:\ 2^{p-1}\!\equiv 1\equiv 2^n\,$ so $\,2\,$ has order dividing $\,p-1\,$ & $\,n.\,$
But for $\:\!p\:\!$ least, $\,p-1\,$ & $\,n\,$ are coprime, so $\,2\,$ has order $\,1,\,$ so $\,p\mid 2^1-1,\,$ contradiction.
Note $ $ This problem was posted to sci.math on 2009\11\03. $\ $ There I remarked that the proof shows that if $\, a^n = 1,\ a\neq 1\,$ then the order of $\,a\,$ is $\,\ge\,$ the least prime factor $\,p\,$ of $\,n.\,$ In particular this implies that $\, a^{p-1}\!\ne 1,\, $ which settles the problem at hand.

For completeness here is a proof of a slightly more general result.
Theorem $\,\  m,n>1,\,\ m\mid 2^{\large n}-1\,\Rightarrow\, \ell(m) > \ell(n),\ \ $ $\ell(x) =\,$ least prime factor of $\,x $
Proof $\,\ \ \ {\rm mod}\,\ q = \ell(m)\!:\,\ 2^{\large n}\equiv 1\equiv\, 2^{\large\color{}{q-1}}\,$ $\Rightarrow$ $\,\ \ell(m) = q > {\rm ord}\,2\color{#80f}{ \ge \ell(n)}$
Remark $ $ The $ $ key idea $ $ is: $\ \  2^n\equiv 1,\,\ \color{#0a0}{2\not\equiv 1}\,$ implies the order of $\,2\,$ is $\,\color{#80f}{\ge\ {\rm least\  prime}\,\ p\mid n},\ $ because the order  must divide $\,n,\,$ and is $\color{}{\neq \color{#c00}1}$ (else $\,\color{#0a0}{2^{\color{#c00}{\large 1}}\equiv 1}),\,$ and the least divisor $>1$  of $\,n\,$ is its least prime factor (by existence and uniqueness of prime factorizations).
In this answer we prove the following generalization
Lemma $ $ If $\,\gcd(b,c)\!=\!1\ $ & $\ 1<n\mid b^n\!-c^n$ then $\,p\mid b\!-\!c\,$ for $\,\color{c00}{p}=$ least prime factor of $\,n$
A: You can look at this modulo the smallest prime factor $p$ of $n$.
Write $n=p^rm$ where $m$ is a product of primes greater than $p$. Then $m$ will be coprime to $p-1$, so $ma\equiv1\pmod{p-1}$ for some positive integer $a$. Now use Fermat's little theorem
$$
(2^n)^a=2^{p^rma}\equiv2^{p^r}\equiv2\not\equiv 1\pmod p.
$$
[Edit: Even better - $n$ will be coprime to $p-1$, as Bill notes in his answer, so there is no need to extract out the $p^r$ factor.]
A: Let $n>1$ and $q$ a prime divisor of $2^n-1$ (note that $q$ is odd). 
We have 
$q \mid 2^n-1$, and also $q\mid 2^{q-1}-1$ (Fermat's little theorem), and therefore
$$q \mid 2^{(q-1, n)}-1.$$
(This follows by considering the order of the element $2$ in $(\mathbb{Z}/q)^{\star}$, or from the fact that $(x^a-1,x^b-1)=x^{(a,b)}-1$.)
In particular, $q\mid 2^n-1$ implies $(q-1,n)>1$. Therefore, any prime factor $q$ of $2^n-1$ is $\equiv 1 \pmod p$ for some prime divisor $p$ of $n$. (Also true for prime factors of $(k+1)^n - k^n$.)
Take for example $2^{21}-1=7^2\cdot 127 \cdot 337$. 
Therefore, the smallest prime divisor $p_0$ of $n$ cannot divide $2^n-1$, and so neither can $n$.
A: We only need to prove for positive $n$, as for negative $n$, $2^n - 1$ is not an integer hence not divisible by $n$. $n = 0$ is meaningless for this proof.
Let $p$ be the least prime divisor of $n$. Let $n = pk$, $k \in Z^+$
$p$ is co-prime to $p-1$ , and $k$ is a product of primes greater than or equal to $p$ which are obviously co-prime to $p-1$ (as all numbers less than a prime are co-prime to it).
This means that $n$ is coprime to and greater than $p-1$, and using the division algorithm can be expressed as:
$n = q(p-1) + r$ where $r \lt p-1, r \neq 0$
This means that $2^n$ can be rewritten as:
$2^n = 2^{q(p-1) + r} = 2^{q(p-1)}*2^r$
By Fermat's Little Theorem, we know that $2^{(p-1)} \equiv 1 \mod p$.
So we get:
$2^n = 2^{q(p-1)}*2^r \equiv 1^q*2^r \equiv 2^r\mod{p}$
However, we know that $r \lt p - 1$. Additionally, $\gcd(r,p-1) = 1$. We know this because if $\gcd(r,p-1) = d > 1$, then $d \mid q(p-1) + r $, so $d \mid n$. But we know that $\gcd(p-1,n)$ = 1 so that's a contradiction. So $\gcd(r,p-1) = 1$.
Let $w$ be the least number that satisfies the congruence $2^w \equiv 1 \mod{p}$.
In other words $w$ is the order of 2 mod p.
We know that $w$ divides $p-1$, and $w$ cannot divide $r$ as $\gcd(p-1,r) = 1$.
If you're not familiar with the concept of order mod p, you can actually prove pretty easily that if $w$ is the least number such that $2^w \equiv 1 \mod{p}$, $w$ must divide any number $i$ which satisfies the congruence $2^i \equiv 1 \mod{p}$.
As $w$ must divide any $i$ for which $2^i \equiv 1 \mod p$, and $w \nmid r$, we can conclude $2^r \not\equiv 1 \mod p \implies 2^n \not\equiv 1 \mod p$, so p does not divide $2^n - 1$.
If the least prime factor of n doesn't divide $2^n - 1$, then it follows that n doesn't.
