Let $\displaystyle N$ be the set of all nilpotent elements of a commutative ring $\displaystyle R$, and show that $\displaystyle N\subseteq P$ for each prime ideal $\displaystyle P$ of $\displaystyle R$.

*Proof:* Let $\displaystyle p\in P$ and $\displaystyle a\in N$. Then there is a positive integer $\displaystyle m\geq 2$ with $\displaystyle a^m=0$. Assume towards a contradiction that $\displaystyle a\notin P$. Since $\displaystyle a^{m-1}a=a^m=0=0p\in P$ and $\displaystyle a\notin P$, then by definition of prime ideals we have $\displaystyle a^{m-1}\in P$. However, the same can be said for any integer $\displaystyle n\geq 2$, that is, if $\displaystyle a^n\in P$ then $\displaystyle a^{n-1}a\in P$ and therefore $\displaystyle a^{n-1}\in P$. By induction, since $\displaystyle a^m\in P$, then $\displaystyle a\in P$, which contradicts our assumption that $\displaystyle a\notin P$. So $\displaystyle a\in P$ whenever $\displaystyle a\in N$, that is, $\displaystyle N\subseteq P$. $\displaystyle \blacksquare$

The reasoning is sound, but I'm not quite sure how to best articulate it. I don't think "by induction" is completely appropriate, here. Also, I like to avoid by-contradiction proofs if possible. Does anyone have an idea for how I can more elegantly and accurately explain what's going on in this exercise?

Thanks!