I want to make really sure that I am not making a mistake here, although I generally don't have problem with asymptotics.
I have proven that a certain function $f(n)\in o(2^n)$. Well actually I proved it is $O(n^\sqrt{n})\subset o(2^n)$. Which follows since $$\lim_{n\rightarrow\infty}\dfrac{n^\sqrt{n}}{2^n}=0$$
Based on my understanding of little o, it is submultiplicative(i.e. $o(f)o(g)\subset o(fg)$). This implies then that $o(2^n)\subset 2^{o(n)}$. Or am I making a grave mistake here? Can I get the bound I am looking for another way?
Suppose $\epsilon(n)=o(n)$ but $\epsilon(n)\to+\infty$ for a function $\epsilon$. Then $2^{n-\epsilon(n)}$ is $o(2^n)$ but not $2^{o(n)}$.
For the desired bound: Suppose $f(n)$ is $O(n^{\sqrt{n}})$, equivalently $O(2^{\sqrt{n}\log_2n})$ with constant $C$. Then $f(n)\le C2^{\sqrt{n}\log_2n}\implies \log_2 f(n)=C'+\sqrt{n}\log_2n$ is $o(n)$ hence $f(n)$ is $2^{o(n)}$ ($f>0$ assumed).