Prove that there exists a natural number n for which $11\mid (2^{n} - 1)$ I'm thinking putting it into modulo form: there exists a natural number $n$ for which
$$2^{n}\equiv 1 \pmod {11}$$
but I don't know what to do next and I'm still confused how to figure out remainders when doing modulos, like $2^n\equiv \;?? \pmod{11}$. Is there some pattern to find $??$ or you would have to use specific numbers for $??$ which is divisible by $11$?

A natural number $n$ will have the property that $11\mid 2^n-1$ precisely when $n$ is a multiple of $10$.
$$\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
n & \!\!\!\!\!& 1 & 2& 3& 4& 5& 6& 7&8&9&\mathbf{\Large 10}&11&12&13&14\\\hline\\
2^n\bmod 11 & \!\!\!\!\!& 2 & 4& 8 &5 &10 &9 & 7&3 &6&\mathbf{\Large 1 }&2&4&8&5
\end{array}\;\;\cdots\;\;\begin{array}{|c|c|}19 &\mathbf{\Large20}\\\hline\\ 6&\mathbf{\Large1}\end{array}\;\;\cdots$$
To be even more explicit,

Here is a proof that there exists a natural number $n$ such that $2^n\equiv 1\bmod 11$. Consider $n=10$: $$2^{10}-1=1024-1=1023=3\times \fbox{11}\times 31$$
  so that $11\mid 2^{10}-1$. Thus by definition $2^{10}-1\equiv0\bmod 11$, and therefore $2^{10}\equiv 1\bmod 11$.

and 

Here is a proof that there exists a natural number $n$ such that $2^n\equiv 1\bmod 11$. Consider $n=20$:  $$2^{20}-1=1,048,576-1=1,048,575=3\times 5^2\times \fbox{11}\times 31\times 41$$
  so that $11\mid 2^{20}-1$. Thus by definition $2^{20}-1\equiv0\bmod 11$, and therefore $2^{20}\equiv 1\bmod 11$.

