Big - Oh proof $n^{2^n} = O(2^{2^n})$

Question

But the book asks me to prove that it's correct: $$n^{2^n} + 6*2^n = O(2^{2^n})$$ But I think, it's an incorrect one. Because, it's correct only for $n < 2$.

Answer 1

$$f(x)=O(g(x))\quad\mbox{ iff }\quad\limsup_{x\rightarrow\infty}\frac{f(x)}{g(x)}<\infty$$

But

$$\limsup_{n\rightarrow\infty}\frac{n^{2^n}}{2^{2^n}}=\limsup_{n\rightarrow\infty}\left(\frac{n}{2}\right)^{2^n}=\infty$$

Hence $n^{2^n}\neq O(2^{2^n})$