smc

From (ii), note that \begin{alignat}{8} f(2) &= 1\cdot f(1) &&= 1, \\ f\left(2^2\right) &= 2\cdot f(2) &&= 2, \\ f\left(2^3\right) &= 2^2\cdot f\left(2^2\right) &&= 2^{2+1}, \\ f\left(2^4\right) &= 2^3\cdot f\left(2^3\right) &&= 2^{3+2+1}, \end{alignat} and so on. In general, we have $$f\left(2^n\right)=2^{(n-1)+(n-2)+(n-3)+\cdots +3+2+1}$$ for any positive integer $n.$ Therefore, the answer is $$\begin{array}{rl}f\left(2^{100}\right)& =2^{99+98+97+\cdots +3+2+1}\\ & =2^{99\cdot 100/2}\\ & =\boxed{2^{4950}}.\end{array}$$