Prove that if a prime $p = 2^k + 1$, it is a Fermat prime

Question

So, I encountered this issue in the proof of the constructability of a regular n-gon, where it is shown that the n-gon is constructable iff it is a product of $2^k$ and distinct primes $p_i$ such that $p_i-1 = 2^k$. I understand that, however, I don't understand how we can follow that if $p - 1 = 2^k$ $p$ must be a Fermat prime, i.e. $p = 2^{2^s} + 1$. It is clear that if $k$ would divide $p-1$, $k$ would be a power of two, and my textbook claims that it is elementary to see that $k$ must divide $p-1$ - however, I don't know how I could prove that. What am I missing?