Is there a function satisfying the following properties $ f^{n}(x)=(f(x))^{n+1}$?? 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 $f^{(0)}(x) = f(x)$.

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 $f(x) \equiv 0$.
