Why do we discard the constant 1 in proving that $2^{n+1} = O(2^n)$
No, there exists such a constant. Indeed take $c=2$, then
$$\forall n \in \mathbb{N}\quad 2^{n+1}=2^1 \cdot 2^n \le  c \cdot 2^n.$$
So $2^{n+1}=O(2^n)$.