I'm having trouble rewriting this in terms of n. I'm trying to compare this with other asymptotic functions.
$$16^{\sqrt{\log_2n}}$$
I have these other ones which I have already ordered: $\log_2 n, (\log_2 \log_2 n)^2, (\log_2 n)^2, \sqrt{n},$ and $2^{\log_3 n}$
I know that for something like $2^{\log_3 n}$, you can rewrite it as $n^{\log_3 2}$
On the assumption that $n \in \mathbb{N}$, we obtain $$ \begin{eqnarray} 16^{\sqrt{\log_2(n)}} &=& n^{\log_{n}(16) \sqrt{\log_2(n)}} \tag{$\small a = b^{\log_b(a)}$} \\ &=& n^{\sqrt{\log_2(n) \cdot \log_{n}(16) \cdot \log_{n}(16)}} \tag{$\small a = \sqrt{a \cdot a}$} \\ &=& n^{\sqrt{\log_2(16) \cdot \log_{n}(16)}} \tag{$\small \log_a(b) = \log_a(c) \cdot \log_c(b)$} \\ &=& n^{\sqrt{4 \cdot \log_{n}(16)}} \\ &=& n^{\sqrt{\log_{n}(65536)}} \end{eqnarray} $$ However, you cannot get rid of the $n$ in the exponent, because of the root.