I found the following question and can't make much headway with it.
Show that there exist infinitely many positive integers $n$ such that $2^n \equiv n \mod p$ where $p$ is a odd prime...
I started by writing the number as $n=(p-1)k+c$ which implies $$2^c \equiv c-k \mod p$$ where I used Fermat's Little theorem. Now if $k$ is of the form $pm$ for some integer $m$ we get $n=(p-1)(p)m+c$ $$2^c \equiv c \mod p$$ and we have reduced $n$ to some smaller integer $c$ but I can't figure out how to find the smallest number $c$ which would then generate infinitely many $n$.
Any help in finding a possible approach to this would be highly appreciated.