Prove that $n\log(n) = O(2^{\log^2(n)})$ I have to prove that $n\log(n) = O(2^{\log^2(n)})$, where $\log(x) = \log_2x$. I tried saying that: $2^{\log^2(n)} = 2^ {\log(n)\log(n)}= 2^{\log(n)^{\log(n)}} = n^{log(n)}$. I also know that $n\log(n) = \log(n^{n})$.I thought of just saying that it's true because the first function is logarithmic, but I am not sure this is correct. I also tried with the fact that $n\log(n) = O(2^n)$, but proving that $2^{\log(n)} = O(2^n)$ doesn't help. Any recommendations would be appreciated.
Note that $\log n + \log \log n \le 2 \log n \le (\log n)^2$ provided that $n \ge 4$ in which case $$n \log n = 2^{\log n + \log \log n} \le 2^{(\log n)^2}.$$
Actually, $n\log n=o(2^{\log^2 n})$ (little oh), which is a stronger statement.
To see the, consider the functions $f(x)=x\log x$ and $g(x)=2^{\log^2 x}$.
Clearly both functions tend to infinity as $x\to\infty$. Using L'hopital's rule,
$$\lim_{x\to\infty}\frac{f(x)}{g(x)}=\lim_{x\to\infty}\frac{f'(x)}{g'(x)}=\frac{
\frac{1+\ln x}{\ln 2}
}
{\frac{2\exp(\frac{\ln^2 x}{\ln 2})\ln x}{x\ln 2}}=\frac{x\ln 2(1+\ln x)}{2\ln 2\exp(\frac{\ln^2 x}{\ln 2})\ln x}$$
Since for sufficiently large $x$ we have $\ln x&gt;2\ln 2$, we deduce that
for sufficiently large $x$ $\exp(\frac{\ln^2 x}{\ln 2})&gt;\exp(2\ln x)=x^2$, and so the r.h.s of the displayed equation above is bounded above by
$$\frac{x\ln 2(1+\ln x)}{2\ln 2\cdot x^2\ln x}=O(\frac{1}{x})$$
Therefore $f(x)=o(g(x))$ as $x\to\infty$, and in particular, $n\log n=o(2^{\log^2 n})$ as $n\to\infty$.
Remark. The argument above shows that actually a much stronger statement is true, namely for every positive integer $k$,
$$\frac{f(x)}{g(x)}=o(x^{-k})\quad (x\to\infty)$$
Therefore, the ratio
$$\frac{n\log n}{2^{\log^2 n}}$$
tends to zero faster than any power of $\frac{1}{n}$ as $n\to\infty$.
In other words, for every positive integer $k$,
$$\lim_{n\to\infty}\frac{n^k\log n}{2^{\log^2 n}}$$
This of course answers the OP's question, although its  primary purpose is to show that the OP's question, is in a sense, &quot;not the right question&quot;.