Let $f:\mathbb{R}\to (0,\infty)$ be a differentiable function satisfying $$f(f(x))=f^\prime(x)$$for each $x$. Show no such function exists.
I got this problem in an exam. I haven't done anything significant with it. I have found that $f^\prime=f\circ f>0$ so $f(f(x))>f(0)$ hence we have $f^\prime(x)>f(0)$. But I have no idea how to use it. I tried to apply the mean value theorem on $$\frac{f(f(x))-f(0)}{f(x)}=f^\prime(c)=\frac{f^\prime(x)-f(0)}{f(x)}$$ but that doesn't lead anywhere. Can someone help me? Thanks a lot.