= Solution
A <Noetherian ring> is a ring whose <ideals> satisfy the <ascending chain condition>; equivalently every <ideal> is finitely generated. For the surjective <ring endomorphism>, put $K_j=\ker f^j$. The ascending chain of <ideals> $K_1\subseteq K_2\subseteq\cdots$ stabilizes, say $K_m=K_{m+1}$. If $a\in\ker f$, surjectivity of $f^m$ gives $b$ with $f^m(b)=a$. Then $f^{m+1}(b)=0$, so $b\in K_{m+1}=K_m$ and $a=0$. \b[The <endomorphism> is therefore injective], hence a <ring automorphism>.
The <nilradical> is the <ideal>
$$
N(R)=\sqrt{(0)}=\{a\in R:a^j=0\text{ for some }j\geq1\}.
$$
For example, if $a^u=b^v=0$, the binomial expansion gives $(a+b)^{u+v-1}=0$; multiplication by any ring element also preserves nilpotence. Every <prime ideal> contains each <nilpotent element>, since $a^j\in P$ implies $a\in P$. Conversely, if $a$ is not nilpotent, the <multiplicative subset> $\{1,a,a^2,\ldots\}$ avoids $(0)$. By <Zorn's lemma>, choose an <ideal> maximal among those disjoint from this subset. It is a <prime ideal>: if $xy\in J$ but neither factor is in $J$, then $J+(x)$ and $J+(y)$ both meet the subset, and multiplying their two witnesses puts a power of $a$ in $J$, a contradiction. This <prime ideal> excludes $a$. \b[Hence]
$$
\boxed{N(R)=\bigcap_{P\in\operatorname{Spec}R}P.}
$$
For the zero ring the empty intersection is the whole ring, as required.
Here is a <finite prime-intersection representation of a radical> argument that does not assume the result. In a <Noetherian ring>, suppose some <ideal> $J$ has <ideal radical> not expressible as a finite intersection of <prime ideals>, and choose such a $J$ maximal by the <ascending chain condition>. It is proper and cannot be prime. Choose $a,b\notin J$ with $ab\in J$. The two larger <ideals> $J+(a)$ and $J+(b)$ have finite prime-intersection radicals, and
$$
\sqrt J=\sqrt{J+(a)}\cap\sqrt{J+(b)}.
$$
Indeed, if $x^u\in J+(a)$ and $x^v\in J+(b)$, then $x^{u+v}\in J+(ab)=J$. This contradicts the choice of $J$. Applying the result to $(0)$ proves \b[that $N(R)$ is the intersection of finitely many <prime ideals>]. Empty intersections handle the unit <ideal>.
For <reducedness of a formal power series ring>, the inclusion of constants gives one implication: a nonzero <nilpotent element> of $R$ remains nonzero and nilpotent in $R[[X]]$. For the other, let $R$ be a <reduced ring> and take a nonzero <formal power series> $F=\sum_{j\geq r}a_jX^j$ with $a_r\ne0$. The coefficient of $X^{mr}$ in $F^m$ is $a_r^m\ne0$, because $R$ is reduced. Thus $F$ is not nilpotent. \b[Therefore]
$$
\boxed{N(R)=0\quad\Longleftrightarrow\quad N(R[[X]])=0.}
$$
No <Noetherian ring> hypothesis is needed for this argument.
For a <nilradical that is not nilpotent>, take
$$
\boxed{R=k[t_1,t_2,\ldots]/(t_1^2,t_2^2,\ldots),\qquad N(R)=(\bar t_1,\bar t_2,\ldots).}
$$
Every element of this <ideal> uses finitely many variables and is nilpotent: an element using $s$ variables lies in an <ideal> whose $(s+1)$st power is zero. The quotient by this <ideal> is the <field> $k$, so it is exactly the <nilradical>. Nevertheless $\bar t_1\cdots\bar t_m\ne0$ for every $m$, because the square-free <monomials> form a <vector space basis>. Thus $N(R)^m\ne0$ for all $m$.
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