Show that $17$ divides $p-1$ I'm given that $s=2^{17}-1$ and that $p$ is a prime factor of $s$.
First I'm asked to show that $2^{17}\equiv 1(\mod p).$
For this I have simply said that since $p$ divides $s$, this means that $2^{17}-1=pk$ and so $2^{17}\equiv 1(\mod p).$
Next I'm asked to show that $17$ divides $p-1.$
I'm think I'm expected to be using Fermat's Little Theorem here, I'm not getting anywhere with it at the moment though. Is this the correct approach and how would one do it?

$2^{17}\equiv 1\mod p$ says the order of $2$ modulo $p$ is a divisor of $17$. However $17$ is prime, and the order of $2$ is not $1$. Hence this order is $17$. From Little Fermat there results $17$   divides $p-1$.
