$f(n) = 2^{n+1} = O(2^n)$
Intuitively, I think the statement is false. However, when I go about disproving it, I find that $2^{n+1} = 2^n \cdot 2$, meaning that if there is a constant $C$ larger than $2$, then $|2^{n+1}| \le C|(2^n)|$, thus proving the statement.
Any hints would be greatly appreciated.
You have successfully proved that $2^{n+1} \in O(2^n)$ with the implicit constant $2$.