Is there any function which grows 'slower' than its derivative?

Does a function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f'(x) > f(x) > 0$ exist?

Intuitively, I think it can't exist.

I've tried finding the answer using the definition of derivative:

1. I know that if $\lim_{x \rightarrow k} f(x)$ exists and is finite, then $\lim_{x \rightarrow k} f(x) = \lim_{x \rightarrow k^+} f(x) = \lim_{x \rightarrow k^-} f(x)$

2. Thanks to this property, I can write:

$$\begin{align} & f'(x) > f(x) > 0 \\ & \lim_{h \rightarrow 0^+} \frac{f(x + h) - f(x)}h > f(x) > 0 \\ & \lim_{h \rightarrow 0^+} f(x + h) - f(x) > h f(x) > 0 \\ & \lim_{h \rightarrow 0^+} f(x + h) > (h + 1) f(x) > f(x) \\ & \lim_{h \rightarrow 0^+} \frac{f(x + h)}{f(x)} > h + 1 > 1 \end{align}$$

3. This leads to the result $1 > 1 > 1$ (or $0 > 0 > 0$ if you stop earlier), which is false.

However I guess I made serious mistakes with my proof. I think I've used limits the wrong way. What do you think?

expanded from David's comment

$f' > f$ means $f'/f > 1$ so $(\log f)' > 1$. Why not take $\log f > x$, say $\log f = 2x$, or $f = e^{2x}$.

Thus $f' > f > 0$ since $2e^{2x} > e^{2x} > 0$.

added: Is there a sub-exponential solution?
From $(\log f)'>1$ we get $$ \log(f(x))-\log(f(0)) > \int_0^x\;1\;dt = x $$ so $$ \frac{f(x)}{f(0)} > e^x $$ and thus $$ f(x) > C e^x $$ for some constant $C$ ... it is not sub-exponential.

There are multiple mistakes in your proof (i.e. dividing by $f(x)$ is not necessarily okay, since it is not necessarily positive). The most major is that you treat the variable in the limit as if it were not "bound". That is, if you have something like $$\lim_{h\rightarrow 0^+}1>0$$ which is true, you can't necessarily, say, multiply through by $h$ to get $$\lim_{h\rightarrow 0^+}h>0\cdot h$$ which is false. This is essentially what you do, and why your conclusion is wrong. You have to regard the $h$ as belonging to the limit - that is $\lim_{h\rightarrow 0^{+}}f(h)$ is a constant - it does not depend on $h$, because there is no notion of "$h$" outside of the limit.

For an example of a function that does not satisfy this, you can take $e^{\alpha x}$ for any $\alpha>1$. Another solution is $xe^x$. It's worth noting that since the solution to $f'=f$ is $x\mapsto e^x$ we can prove that any solution grows faster than exponentially.