Function Game

Use $\color{#D61F06}{\small{f(2n)=nf(n)}}$ repeatedly: $\begin{aligned}f(2^n)=f(2\cdot 2^{n-1})=&2^{n-1}f(2^{n-1})\\=&2^{n-1} 2^{n-2}f(2^{n-2})\\=&2^{n-1}2^{n-2}2^{n-3}f(2^{n-3})\\&\cdots\\&\cdots\\=&2^{(n-1)+(n-2)+(n-3)+\cdots +1}\underbrace{f(1)}_{\color{#3D99F6}{1}}\\=& 2^{\dfrac{n(n-1)}2}\end{aligned}$

Hence , $f(2^n)=2^{\dfrac{n(n-1)}2}$ $f(2^{100})=2^{\dfrac{100(100-1)}2}=\boxed{2^{4950}}$

From the first few $f(2^k)$ , where $k$ is non-negative integer, we notice that $f(2^k) = 2^{\frac {k(k-1)}2}$ . Let us prove that the claim $f(2^k) = 2^{\frac {k(k-1)}2}$ is true for all $k$ .

1. For $k=0$ , $f(2^0) = f(1) = 2^{\frac {0(0-1)}2} = 1$ , as given, therefore, the claim is true for $k=0$ .

2. Assuming the claim is true for $k$ , then:

$\begin{aligned} \quad \quad f(2^{k+1}) & = f(2\cdot 2^k) = 2^kf(2^k) = 2^k \cdot 2^{\frac {k(k-1)}2} = 2^{\frac {k(k+1)}2} = 2^{\frac {(\color{#3D99F6}{k+1})(\color{#3D99F6}{k+1}-1)}2} \end{aligned}$

$\quad$ Therefore, the claim is also true for $k+1$ and hence all $k$ .

Therefore, $f(2^{100}) = 2^{\frac {100\times 99}2} = \boxed{2^{4950}}$ .