Let $f:\mathbb{R} \to \mathbb{R}$ continuous with $f(f(x))=e^x$, show that $\lim_{x\to \infty } \frac{f(x)}{x^n}=\infty$ (Brazilian Olympiad) Given $f:\mathbb{R} \to \mathbb{R}$ continuous with $f(f(x))=e^x$ for all $x\in\mathbb{R}$. Show that $\lim_{x\to \infty } \frac{f(x)}{x^n}=\infty$ for all $n\in\mathbb{N}$.
I appreciate any help!
Edit 1: It is easy to see that $f$ is injective. So, if $f$ increases anywhere, $f$ will be an increasing function.
Edit 2: As said in the comments, $f$ must indeed be increasing, since
$f(1)=f(e^0)=f(f(f(0)))=e^{f(0)}>f(0)$.
 A: remark
It is not enough to assume (for example) that $f(f(x)) = e^x$ for all $x>1$.  You must assume (as the OP says) $f(f(x))=e^x$ for all $x$.  This is used in Edit 2.
The following function $f : \mathbb R \to \mathbb R$
is continuous and $f(f(x)) = e^x$ for all $x > 1$ but fails the required
conclusion.

$$
f(x) = \begin{cases}
-2x,\quad &x\ge T
\\
e^{-x/2},\quad &x < T
\end{cases}
$$
Here $T \approx -0.7148$ is one of the points where the two graphs $-2x,e^{-x/2}$ cross.
A: We'll start with a weak estimate, and then use the functional equation to improve it iteratively.

Claim: There is a constant $d > 0$ such that $ f(x) \ge x+d$ for all $x \ge 0$.

Once we have that estimate (see below for the proof), we can can continue as follows: Setting $C = e^d > 1$ we get
$$
f(x) = f(e^{\ln x}) = e^{f(\ln x)} \ge e^{\ln x + d} = xe^d = Cx
$$
for $x \ge 1$, then
$$
f(x) = e^{f(\ln x)} \ge e^{C\ln x} = x^C
$$
for $x \ge e$, and finally
$$
f(x) = e^{f(\ln x)} \ge e^{(\ln x)^C} 
$$
for $x \ge e^e$.
It follows that
$$
\frac{f(e^x)}{e^{nx}} \ge \frac{e^{x^C}}{e^{nx}} = e^{x^C-nx} \ge x^C-nx+1
$$
and therefore
$$
 \lim_{y \to \infty} \frac{f(y)}{y^n} = \lim_{x \to \infty} \frac{f(e^x)}{e^{nx}} = +\infty \, .
$$

Proof of the claim: $f(0) > 0$ and $f$ has no fixed points (see for example thoughts about $f(f(x))=e^x$), therefore $f(x) > x$ for all $x \in \Bbb R$. Then
$$
 d = \min \{ f(x) - x \mid 0 \le x \le 1 \}
$$
is strictly positive. Define the sequence $(E_k)$ recursively by $E_0 = 0$, $E_{k+1} = e^{E_k}$:
$$
 0, 1, e, e^e, e^{e^e},\ldots
$$
$f(x) \ge x+d$ holds for $E_0 \le x \le E_1$, and if it holds for $E_k \le x \le E_{k+1}$ then
$$
 f(x) =  e^{f(\ln x)} \ge e^{\ln x + d} = xe^d \\
 \ge x(1+d) = x + xd \ge x+d
$$
for $E_{k+1} \le x \le E_{k+2}$. This concludes the proof of the claim, since $E_k \to \infty$.
