Showing a function on $[0,1]$ is the zero function Suppose $f$ is continuous on $[0,1]$ and 
$$ \int\limits_0^1 f(x) x^n d x = 0 $$ 
for all $n=0,1,2,...$. Then, $f(x) = 0$
I was trying to use by parts
$$ 0 = f(x) \frac{x^{n+1}}{n+1} \bigg|_0^1 - \int\limits_0^1 \frac{x^{n+1}}{n+1} f(x) dx $$
Thus,
$$ \frac{f(1)}{n+1} = \int\limits_0^1 \frac{ x^{n+1} f(x) dx }{n+1} \implies f(1) = \int\limits_0^1 x^{n+1} f(x) dx $$
but here I am stuck. Maybe my idea is not the best one? How would you approach this problem?

Since $C^0([0,1])\subset L^2(0,1)$ we may assume that $f(x)$ has the following expansion in terms of the shifted Legendre polynomials (forming an orthogonal base of $L^2(0,1)$ with respect to the usual inner product $\langle f,g\rangle=\int_{0}^{1}f(x)g(x)\,dx$)
$$ f(x) = \sum_{n\geq 0} c_n\,P_n(2x-1) \tag{1}$$
where
$$ c_n = (2n+1)\int_{0}^{1} f(x)\,P_n(2x-1)\,dx = 0 \tag{2} $$
since $P_n(2x-1)$ is a polynomial. $f(x)\equiv 0$ then follows from Parseval's identity:
$$ \int_{0}^{1}f(x)^2\,dx = \sum_{n\geq 0}\frac{c_n^2}{2n+1}=0.\tag{3}$$
