If $\int_0^1 f(x) e^{nx} dx = 0$ for every n, then f=0 $f$ be a continuous function [0,1] to $R$.
$\int_0^1 f(x)e^{nx} dx = 0$ for all $n \in N\cup\{0\}$
how to prove $f(x)= 0$ in $[0,1]$ for all $x\in[0,1]$?
I solved "$\int_0^1 f(x)x^n dx = 0$ for all $n \in N\cup\{0\}$" with weierstrass theorem

You haved showed that if $\int_a^b f(x)x^n dx = 0$, then $f=0$. well, we will use it. 
Let $e^x=y$, 
$$\int_0^1 f(x)e^{nx}\, \mathrm dx=\int_1^e f(\ln y)y^{n-1}\,\mathrm dy$$
so  $\int_1^e f(\ln x)x^{n-1}\, \mathrm dx=0$ for all $n \gt 0$. Hence
$$f(\ln x)=0, x\in [1,e]$$
that is $f(x)=0, x\in [0,1]$
