I would like to show that $n^{log n} = o(2^n)$.
Here is my attempt:
I see that $\log{(n^{\log{n}})} = (\log{n})^2,$ and $\log{2^n} = n\log{2}$. I also know that $(\log{n})^2=o(n)$, so that for any $d>0$ eventually $(\log{n})^2 < d{n}.$
So, one approach would be to let $c>0$ and try to work backwards from $n^{\log{n}} < c2^n \rightarrow (\log{n})^2 < \log{c}+n\log{2}$, and figure out what $d$ I need...but this seems unlikely since $\log{c}$ can be arbitrarily negatively large.
I also tried looking at it more generally...showing that if $\log{g(n)} = o(\log{f(n)})$, then it must be that $g(n) = o(f(n))$. But I run into the same issue as in the above concrete example.
It isn't to hard to show that $n^{\log n} = O(2^n)$. In fact there is a constant $M$ satisfying $0 \le (\log x)^2 \le M + x \log 2$ for all $x > 1$.
Consider the following application of L'Hospital's rule: since $\displaystyle \lim_{x \to \infty} \frac{x \log x}{2^x}$ takes the form $\dfrac{\infty}{\infty}$ you have $$\lim_{x \to \infty} \frac{x \log x}{2^x} = \lim_{x \to \infty} \frac{x \log x \cdot 2 \log x \cdot \frac 1x}{2^x \log 2}$$ provided that the latter limit exists. But since $\dfrac{x \log x}{2^x}$ is bounded above and below (provided say $x > 1$) and $\dfrac{\log x}{x} \to 0$, you may conclude that $$\lim_{x \to \infty} \frac{x \log x \cdot 2 \log x \cdot \frac 1x}{2^x \log 2} = 0.$$