Show that the ring $C^\infty(X, \mathbb{R})$ is a reduced ring. I've already shown that $C^\infty(X, \mathbb{R})$ is not an integral domain, since the product of two nonzero functions $fg$ can be zero as long as the intersection of their supports is empty. Now I'm looking to show that the ring $C^\infty(X, \mathbb{R})$ is a reduced ring (i.e., it has no nonzero nilpotent elements). This seems more difficult, and I don't see where the smoothness condition comes into play. 
 A: To summarize and expand on the comments:
You've seen now that the ring multiplication (which is pointwise multplication of the functions) essentially passes the question about the function (if $f^n=0$ is $f=0$?) down to the images of points (if $f(x)^n=0$, then is $f(x)=0$?). We saw that since the images of $f$ are in a field, the answer to the latter question is "yes" and then so is the answer to the former question.
Moreover, it's now obvious that $\Bbb R$ wasn't special at all, you just need at least a reduced ring $S$ (with some topology, if you are going to ask questions about continuous functions). Additionally, you saw that smoothness didn't come into play, and you can easily see that the same reasoning works for "the ring of (blah) functions from $X$ into $\Bbb R$" where "blah" is "all/continuous/smooth" or some other thing that makes sense in the context.
Finally, I wanted to point out that while "no nonzero nilpotent elements" is a fine characterization of reduced rings, that is also equivalent to "$\forall x(x^2=0\implies x=0)$". I know it's not a huge change, but it does switch an $n$ into a $2$, and that seems more confidence building :) You might try to prove that equivalence, if you haven't already.
A: Reduced rings are closed under products and subrings - this is obvious. But $C(X,\mathbb{R})$ is a subring of $\prod_{x \in X} \mathbb{R}$.
