I want to check if $2^{2^n}=O(2^n)$.
That's what I have tried:
Let $4^n=O(2^n)$.
Then, $\exists c_1>0, n_1 \in \mathbb{N}$ such that $\forall n \geq n_1$:
$$4^n \leq c \cdot 2^n$$ $$$$
$$\lim_{n \to +\infty} \frac{4^n}{2^n}=\lim_{n \to +\infty} \left( \frac{4}{2}\right)^n=\lim_{n \to +\infty} 2^n=+\infty$$
This means that $2^n=o(4^n)$, i.e. $\forall c>0, \exists n_0 \in \mathbb{N}$ such that $\forall n \geq n_0$:
$$2^n<c \cdot 4^n$$
So, we conclude that we cannot find a $c_1>0$ such that $4^n \leq c \cdot 2^n$.
Therefore, the equality $2^{2^n}=O(2^n)$ does not hold.
Could you tell me if it is right or if I have done something wrong?