Why do we discard the constant 1 in proving that $2^{n+1} = O(2^n)$ In this post
someone said:


  $n + 1$ is approximately equal to 1 for $2^{n+1} = O(2^n)$.


I don't understand this point. For example, in this example: 

$$ n + 1 \leq c \cdot n $$ 

for some $c$ and some $n \geq n_0$, the right-hand side will always be at least the left-hand side. 

But here: $$ 2^{n+1} =O(2^n)$$

the left-hand side will always exceed the right-hand side.
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 $2^{n+1}=O(2^n)$.