Solving $y=\prod_{n=1}^{\infty}\frac{d^ny}{dx^n}$ There is the trivial $y=0$, but beyond that, could there be further solutions for $y$ in terms of $x$ such that 
$$y=\prod_{n=1}^{\infty}\frac{d^ny}{dx^n}\mbox{ pointwise}$$
? I posed this problem to myself, so I have no idea. I began with:
$$\ln(y)=\sum_{n=1}^{\infty}\ln\left(\frac{d^ny}{dx^n}\right)\mbox{ pointwise}$$
$$\frac{y'(x)}{y(x)}=\sum_{n=1}^{\infty}\frac{y^{(n+1)}(x)}{y^{(n)}(x)}\mbox{ pointwise}$$
And I've fiddled with various manipulations, but without success. Any ideas?

If:
$$\frac{y'(x)}{y(x)} = \sum_{n = 1}^{\infty}\frac{y^{(n+1)}(x)}{y^{(n)}(x)}$$
Then $\lim_{n \rightarrow \infty} \frac{y^{(n+1)}(x)}{y^{(n)}(x)} = 0$. It follows that, for $n$ large enough, you have $y^{(n+1)}(x) < \frac{1}{2} y^{(n)}(x)$. If this is the case:
$$y(x) = \prod_{n = 1}^{\infty}y^{(n)}(x) = 0$$
Meaning that $y = 0$. This is not quite a complete solution, since there are potential issues with taking the natural log of non-positive numbers and of moving the derivative past the sum (as you did in the top post). But I think you should be able to fill in the gaps.
