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?

By Weierstrass approximation theorem, there exists a sequence of polynomial $\{f_n\}$ such that $f_n \rightarrow f$ uniformly on $[0, 1]$. 
Hence it follows
\begin{align}
\lim_{n\rightarrow\infty} \int^1_0 f_n(x)f(x)\ dx = \int^1_0 f(x)^2\ dx
\end{align}
but we also know that
\begin{align}
 \int^1_0 f_n(x)f(x)\ dx =0
\end{align}
since
\begin{align}
\int^1_0 x^n f(x)\ dx = 0
\end{align}
for all $n \in \{0, 1, 2, \ldots\}$. Thus, it follows
\begin{align}
\int^1_0 f(x)^2\ dx = 0
\end{align}
which means $f(x) \equiv 0$. 
