jmc

Define the sequence $$x_i=\text{the remainder when }{2}^i\text{ is divided by 100}.$$ Then note that $x_{22}=x_2=4$, and thus this sequence repeats every 20 terms from $x_2$ onward. The desired product is $2^{1+2+3+\dots +99+100}=2^{5050}$. If we can find $x_{5050}$, we will then be done. But since $5050=20\cdot 252+10$, we see that $x_{5050}=x_{10}=24$. Thus our answer is $2\cdot 4=\boxed{8}$.