Answer
$$f(x)=\sqrt[n]{x}$$
where $n \geq 1 $
Work Step by Step
$$\int_{0}^{x}f(t)dt=k(f(x))^{n+1}$$
Where $k$ is the factor of proportionality.
Differentiate both sides and using the $FTC$ it follows:
$$f(x)=k(n+1)(f(x))^{n}f'(x)$$
$$1=k(n+1)(f(x))^{n-1}f'(x)$$
$$1=\left(\frac{k(n+1)}{n}(f(x))^{n}\right)'$$
$$\int1 dx=\int\left(\frac{k(n+1)}{n}(f(x))^{n}\right)'dx$$
$$x+c=\frac{k(n+1)}{n}(f(x))^{n}$$
Since $f(0)=0$ it follows:
$$0+c=\frac{k(n+1)}{n}(f(0))^{n} \to c=0$$
so:
$$x=\frac{k(n+1)}{n}(f(x))^{n}$$
Since $f(1)=1$ it follows:
$$1=\frac{k(n+1)}{n}(f(1))^{n} \to 1=\frac{k(n+1)}{n} $$
so:
$$x=1\cdot (f(x))^{n}$$
$$x= (f(x))^{n}$$
$$f(x)=\sqrt[n]{x}$$
where $n \geq 1 $
