Show that $2^n \not \equiv 1 \pmod n$

Question

Show that $2^n \not \equiv 1 \pmod n$ for any $n$. This is obvious for even $n$, and for odd $n$ we get that $2^n \equiv 2^{\phi(n)} \pmod n$. Can I derive a contradiction from this?

Answer 1

Let $n\geq 3$ odd, and suppose that $2^n$ is congruent to $1$ modulo $n$. let $p$ the least prime divisor of $n$. Then we have $2^n$ congruent to $1$ modulo $p$, and if $l_p$ is the order of $2$ modulo $p$, we have then that $l_p$ divide $n$. But as $2^{p-1}$ is $1$ modulo $p$, $l_p$ divide $p-1$. As clearly $l_p\geq 2$, it has a prime divisor $k<p$, and $k$ divide $n$, a contradiction with the definition of $p$.