TITLE: show $p\mid 2^n-n$ for infinitely many $n$
QUESTION [7 upvotes]: Show that $p\mid 2^n-n$ for infinitely many $n$. $p$ is a prime and $n$ is an integer. 

I tried using Fermat's little theorem and got $2^p-p\equiv2\pmod p$ and $2^{p-1}-(p-1)\equiv2\pmod p$. So I can't even find one $n$ that satisfy the condition. 
Any helps and hints appreciated. (even just one $n$ that works for every $p$)

REPLY [8 votes]: Put $$n = (p-1)(kp-1)$$
where $k$ is arbitrary positive integer.
Then $$ 2^n-n = (2^{p-1})^{kp-1}-(p-1)(kp-1)\equiv 1-1 =0 \pmod p$$