Prove that $ 2^n \not \equiv 1 \pmod{n} $ for any $n > 1$. I have proved this in following way.
Assume that  $ 2^n \equiv 1 \pmod{n} $.
that means $n\mid(2^n -1)$.
but by proof by contradiction, for $n=3$ this does not hold and we can say $n \nmid  (2^n-1) $.
Hence our earlier assumption is wrong.
So, by contradiction we can say that $2^n \not \equiv 1 \pmod{n}$  for any $n$.
My question is, Is this the correct formal proof ?
Can I consider this as correct formal proof?
EDIT: If this is not correct, please provide one.
 A: "For any" is an English term that is not well defined in mathematics in the presence of a negation.  Your sentence can be read as "$2^n  \equiv 1 \mod{n}$ is false for all $n \gt 1$" or as "$2^n  \equiv 1 \mod{n}$ is false for some $n \gt 1$.  If the statement were positive it would clearly be "for all $n \gt 1$"  If you read it as "$2^n  \equiv 1 \mod{n}$ is false for some $n \gt 1$ the example you have shown is sufficient.  If you read it as "$2^n  \equiv 1 \mod{n}$ is false for all $n \gt 1$" one example is not sufficient-you need to prove the failure for all $n \gt 1$
A: We have $2^1\equiv 1\pmod{1}$. Not very interesting! We show there are no others.
For suppose the congruence $2^n\equiv 1\pmod{n}$  holds, and $n\gt 1$. Let $p$ be the smallest prime divisor of $n$. 
Then by Fermat's theorem, 
$$2^{p-1}\equiv 1\pmod{p}.$$ 
Also, since $p$ divides $n$, we have 
$$2^n\equiv 1\pmod{p}.$$
Let $d=\gcd(n,p-1)$. If $d\gt 1$ then $n$ has a divisor greater than $1$ but less than $p$, contradicting the choice of $p$.
Thus $d=1$, and therefore there exist integers $x$ and $y$ such that $(p-1)x+ny=1$. Since $2^{p-1} \equiv 1\pmod{p}$ and $2^{n}\equiv 1\pmod p$, we conclude that $2^1=2^{(p-1)x+ny}\equiv 1 \pmod{p}$.  But $2^1\equiv 1\pmod{p}$ is impossible. 
A: Here's another pass at this:
Suppose for some $k$, we have a positive integer $m$ such that $2^k-1 = mk$. Or, $2^k \equiv 1 (m)$
Therefore, in the vector space $\mathbb{F}_2^k$, there are $mk$ non-zero vectors. Partition these into $m$ disjoint sets of $k$ vectors: $S_i$ for $i=1,2, \ldots m$. Choose vectors $v_i \in S_i$ to form $\{v_1, v_2, \ldots v_k\}$. Now for any other vector, $w$, not in this set, the collection $\{v_1, v_2, \ldots v_k\}\cup\{w\}$ is linearly dependent. Therefore, the set spans $\mathbb{F}_2^k$ and thus it is a basis. Therefore, the number of possible bases is $m^k$. But then, $o(GL_k(\mathbb{F}_2)) = (2^k-1)(2^k-2)\ldots(2^k-2^{k-1})$   
Older attempt
Hope this is right. Suppose there is a $k$ such that $2^k - 1 \equiv 0 (k)$. 
That is, $1+2+2^2+ \ldots +2^{k-1} \equiv 0 (k)$, or, adding the two up: 
$2+2^2+\ldots +2^k \equiv 0 (k)$ 
But this can't be since $(2,k)=1$ as, by hypothesis, $2$ is invertible in $\mathbb{Z}/k\mathbb{Z}$
