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Let $R$ be a commutative ring and $P\triangleleft R$ be a prime ideal. Then $Nil(R)\subset P$.

My approach:

Let $r\in R$, $a\in Nil(R)$. Suppose that $r,a\not\in P$, so that $ra\not\in P$ (since $P$ is prime). Since $a\in Nil(R)$, $\exists n\in\mathbb{N}$ such that $a^n=0$, but $0\in P$, so $a^n\in P$. This implies that $(ra)a^{n-1}\in P$. Suppose $ra\not\in P$, then $a^{n-1}\in P$. Then, inductively, $(ra)a\in P$, so $ra\in P$ or $a\in P$, a contradiction. So $r\in P$ or $a\in P$. Also, $a ra\in P$, and $ra = ar\not\in P$, thus $a\in P$.

I think something peculiar is not correct. Would appreciate a review.

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You have got the main idea right but your choice of $r$ is superfluous (also I don't understand what your last line of the proof is supposed to show).

Since $a\in \operatorname{Nil}(R)$, there exists some $n$ such that $a^n=0\in P$, as you correctly noted. Why not conclude inductively from this that $a\in P$?

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    RIght, I didn't need to overcomplicate this simple proof. – 2017-02-13
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    But where is commutativity of $R$ used? Do we actually need it at all for this case? – 2017-02-13
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    @sequence If you don't have commutativity, then the definition of "prime" is changed so that it is not linked so closely with nilpotent elements. It is no longer true that $x^n\in P\implies x\in P$ for a noncommutative prime ideal $P$. There are lots of nilpotent elements in $M_2(\mathbb R)$ but only one of them lies in the only prime ideal (which is $\{0\}$). – 2017-02-13
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    @rschwieb that would be true if $P$ is completely prime, right? – 2017-02-13
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    @Xam Yes, the condition for "commutative prime ideal" is called "completely prime ideal" in noncommutative ring theory. – 2017-02-13
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    @rschwieb but we aren't using commutativity in the proof in which we inductively arrive at the fact that $aa\in P$, so $a\in P$, are we? – 2017-02-13
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    @sequence Since you possibly are interested in the noncommutative case, here is the analogue: the lower nilradical $Nil_\ast(R)$ is defined as the intersection of all prime ideals. It contains all nilpotent ideals. Finally, the lower nilradical is precisely the set of [strongly nilpotent elements](http://math.stackexchange.com/a/873601/29335). Alternatively you could say "the set of nilpotent elements of a ring $R$ is contained in every completely prime ideal. – 2017-02-13
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    @sequence Yes, you are using commutativity because you are using the commutative definition of prime ideals. – 2017-02-13