show $p\mid 2^n-n$ for infinitely many $n$

Question

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 $p$)

Answer 1

Put $$n = (p-1)(kp-1)$$

where $k$ is arbitrary positive integer.

Then $$ 2^n-n = (2^{p-1})^{kp-1}-(p-1)(kp-1)\equiv 1-1 =0 \pmod p$$