Proof involving set of natural numbers and division algorithm? This is the proof that I am trying to work out:
$ ∀n ∈ \mathbb{N}, 11 \nmid (2^n − 1) $
Obviously this is true because any multiple of power of 2 minus 1 will never equal a multiple of 11, or at least I cannot think of any. But I would like to prove this logically; how would I go about doing so?
My attempt thus far:
(Proof by Contradiction) For all $n \in \mathbb{N}$ let us assume that $11 | 2^n-1$. Therefore, $2^n-1=11j$ for some integer j. Thus, $log_2(11j+1)=n$ which is not an integer? Something like that.
Where do I go from here? Should I be using contradiction? 
 A: This is false. Consider for example,
$$
2^{20}-1=1048575=3\cdot 5^2\cdot 11\cdot 31\cdot 41.
$$
A: A small example of this being wrong: $2^{10}-1=1023=93 \cdot 11$. Actually, I did not need to calculate $2^{10}$ to show this. Using Fermat's Little Theorem  ($a^{p-1} - 1 \equiv 0 \ \text{mod}\ p$ whenever $p$ is prime), with $a=2$ and $p=11$, gives the counterexample directly without any need to compute!
A: This is false: $11$ divides $1023=2^{10}-1$. In general, if $a$ and $b$ are relatively prime, then $a^n\equiv 1$ (mod $b$) for some divisor $n$ of $\phi(b)$.
A: You can notice that $2^5=32$ and $32\equiv (-1)\pmod{11}$ so $32^{2k}-1\equiv (-1)^{2k}-1\equiv 0\pmod{11}$ which gives that $2^{10k}-1$ is always a multiple of $11$.
A: $n\mapsto 2^n\pmod{11}$ maps $\Bbb N$ to a finite set so, by Pigeonhole,  is not injective ($1$-$1$), so there are naturals $\, J < J+K\,$ such that $\,{\rm mod}\ 11\!:\,\ 2^{J+K}\equiv 2^J\,$ so $\, \color{#c00}2^J(2^{K}-1)\equiv 0\,$ thus $\bbox[6px,border:1px solid red]{2^K\equiv 1}\,$ because $\color{#c00}2$ is invertible ($2^{-1}\equiv 6$) so cancellable.
