Which of the given options provides the increasing order of asymptotic complexity of functions $f_1, f_2, f_3$ and $f_4$ ? $$f_1(n) = 2^n$$ $$f_2(n) = n^{3/2}$$ $$f_3(n) = n \log_2 n$$ $$f_4(n) = n^{\log_2 n}$$

To compare the growth rates of the given functions, take the logarithm $\log_2$ of each function:

1. $\log_2 f_1(n) = \log_2(2^n) = n$

2. $\log_2 f_2(n) = \log_2(n^{3/2}) = 1.5 \log_2 n$

3. $\log_2 f_3(n) = \log_2(n \log_2 n) = \log_2 n + \log_2 \log_2 n$

4. $\log_2 f_4(n) = \log_2(n^{\log_2 n}) = (\log_2 n)^2$

Comparing the growth of their logarithms as $n \to \infty$:

Therefore, the increasing order of asymptotic complexity is:

Hence, the correct option is (A).