= Solution
A <nilpotent element> belongs to every <prime ideal>: if $a^r=0\in P$, repeated primality forces $a\in P$. Therefore $N\subseteq\bigcap_P P$.
Conversely, if $a\notin N$, none of its positive powers is zero. The <multiplicative subset> $S=\{1,a,a^2,\ldots\}$ is disjoint from $(0)$. Part (b) supplies a <prime ideal> disjoint from $S$, hence not containing $a$. Thus $a$ is outside the intersection, proving the <nilradical as the intersection of prime ideals> identity:
$$
\boxed{N=\bigcap_{P\text{ prime}}P.}
$$
In the zero ring there are no prime ideals; the empty intersection is interpreted as the whole ring, which is also its nilradical. For nonzero rings this argument proves existence of a prime ideal whenever a nonnilpotent element is available.
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