Prove or disprove: $n\log ({2^n}\log ({n^2})) = O({n^2})$

Question

Prove or disprove: $$n\log ({2^n}\log ({n^2})) = O({n^2})$$

What I reached so far is:

$$\eqalign{ & n({\log _2}{2^n} + \log ({n^2})) = \cr & n(\log n + \log ({n^2})) = ? \cr} $$

Answer 1

$n\log(2^n\log(n^2))=n\log(2^n)+n\log\log(n^2)=n^2\log(2)+n\log(2)+n\log\log n$ Note that the last two are negligible compared to $n^2\log 2$ as $n$ grows large. We can choose $\log 2\le k\le \log 2+\frac{\log 2}{e}$ for $\frac{n^2\log(2)+n\log(2)+n\log\log n}{n^2}=k$ for sufficiently large $n$, so $n\log(2^n\log(n^2))=O(n^2)$

Answer 2

$$n\log(2^n\log(n^2)) = n\log(2^n) + n\log\log(n^2) = n^2\log(2) + n\log(2\log n)$$

For simplicity I assume that $\log = \log_2$. Then this becomes

$$n^2 + n + n\log\log n$$

Is this $O(n^2)$?

Answer 3

Another approach:

$$\frac{\color{red}n\log(2^n\log n^2)}{n^{\color{red}2}}=\frac{n\log2+\log2+\log\log n}n=\log 2+2\frac{\log2+\log\log n}n\xrightarrow[n\to\infty]{}\log2$$