Show that $n^{2^p} \equiv n^2 \pmod {2^p− 1}$ where $2^p− 1$ is prime and it does not divide $n$.
Since $2^p-1$ is prime we can use Corollary 4.2
$$ n^{2^p-1} \equiv n \pmod {2p-1}$$
Multiplying by $n$ gives us the desired conclusion
$$ n^{2p} \equiv n^2 \pmod {2p-1}$$