Answer: D

Solution:

Applying the rule repeatedly, $$\begin{aligned} a_{2^1} &= 2^0, \\ a_{2^2} &= 2^1, \\ a_{2^3} &= 2^{1+2}, \\ a_{2^4} &= 2^{1+2+3}, \ldots \end{aligned}$$ so in general $a_{2^n} = 2^{1 + 2 + \cdots + (n - 1)} = 2^{n(n-1)/2}.$

For $n = 100,$ the exponent is $\dfrac{100 \cdot 99}{2} = 4950,$ so $a_{2^{100}} = 2^{4950}.$

Thus, the correct answer is D.