Using the following definition of radical ideal:
$I$ is radical if and only if $I = \{r \in R \mid \exists n>0: r^n \in I \}$
I have found a direct proof by induction (on Math.SE or here, here) and a direct proof using sub-ideals (see here or here) of this result. Obviously the result is trivial if one uses the (equivalent) definition of radical ideal as the intersection of all prime ideals containing the ideal.
Question: However, the easiest way to prove this result seems to be by contraposition. However, I have not seen anyone use this proof (it admittedly is similar to, but, I think, still simpler than, the induction proof), so I wanted to double-check that it is correct.
The desired result is equivalent to $I$ is not radical $\implies$ $I$ is not prime. Let $I$ not be radical. Then there exists a $P \in R$ such that $P \not\in I$ but $P^m \in I$ for some $m \ge 2$.
Let $m_0$ be the smallest integer such that $P^{m_0} \in I$ (which exists by the well-ordering principle).
Thus since $(P)(P^{m_0 - 1}) =P^{m_0}\in I$, but $P \not\in I$ and $P^{m_0 -1}\not\in I$, $I$ is not prime. $\square$
Looks good. But the induction proof is just as simple (and direct proofs are IMHO a little nicer) :)