Prime radical that is nil but not nilpotent

Question

Please help me to show that the prime radical of the ring $R=\prod\limits_{n = 1}^\infty { \mathbb{Z} /2^n\mathbb{Z} } $ is nil but not nilpotent.

Answer 1

If $\sqrt{0}^n=0$, then in particular $(2 e_n)^{n-1}=0$, i.e. $2^{n-1} = 0 \bmod 2^n$, a contradiction.