Is there a function such that $f' = f\circ f$?

Is there a function $f:\mathbb{R}\rightarrow (0,\infty)$, such that $f' = f\circ f$?

Apparently, I should assume by contradiction there is, and then it should imply that $f$ is increasing but I can't see the reason for that.

EDIT:

Now, we know that $f(0)$ is a lower bound for $f'(x).\forall x \in \mathbb{R}$.
The next claim is for $x<0.f(x) <f(0) + xf(0) = (1+x)f(0)$.

Why it the last claim true?

To answer your specific question: try the fundamental theorem of calculus. Since $f$ is increasing, you know that $f\circ f$ is also increasing, and therefore $f'$ is as well. Therefore, fixing $x$, $f'(x)$ is larger than $f'(y)$ for any $y\lt x$ and since $f'(y) \lt f'(x)$, $\int_0^xf'(y)dy\lt\int_0^xf'(x)dy$ - but the former is $f(x)-f(0)$ and the latter is $xf'(x)$. This gives $f(x)-f(0)\lt xf'(x)$, or $f(x)\lt f(0)+xf'(x)\lt f(0)+xf(0)$, with the last inequality coming by the already-proved result on $f'(x)$.

Observe $f'$ is strictly positive, and so $f$ is monotonically increasing. Thus $f'$ is a monotonically increasing function as well, so we can continuously extend $f$ and $f'$ to $-\infty$, and furthermore we must have $f'(-\infty) = 0$.

Then,

$$0 = \lim_{x \to -\infty} f'(x) = \lim_{x \to -\infty} f(f(x)) = f(\lim_{x \to -\infty} f(x)) $$

Since $f(x) > 0$ for all real $x$, we must have

$$\lim_{x \to -\infty} f(x) = -\infty $$

which is absurd.