Is there a function satisfying the following properties $ f^{n}(x)=(f(x))^{n+1}$??

Question

Is there a function with the following properties?

$$ f(x)=f(x) $$ $$ f'(x)=f(x)^2 $$

$$f^{(n)}(x)=\left(f(x)\right)^{n+1}$$ where $f^{(n)}$ denotes the $n$th derivative, and by convention $f^{(0)}(x) = f(x)$.

Answer 1

We can use the chain rule to show that there is only one such function: $$ f'(x) = f(x)^2, $$ and differentiate both sides and use $f''=f^3$: $$ f(x)^3 = f''(x) = 2f'(x)f(x) = 2f(x)^2f(x), $$ so $f(x)^3=2f(x)^3$, so $f(x) \equiv 0$.

Answer 2

The function $x \mapsto 0: \mathbb{R} \to \mathbb{R}$ is one such.