Proving that $\int_0^1 f(x)e^{nx}\,{\rm d}x = 0$ for all $n\in\mathbb{N}_0$ implies $f(x) = 0$
Solution 1:
Make a change of variables $u = e^x$
$$\int_0^1 f(x) e^{nx}dx = \int_{1}^e g(u) u^{n}du = 0$$
where $g(u) = \frac{f(\log(u))}{u}$ is a continuous function. You can now apply Weierstrass approximation theorem. Since the above holds for all $n$ we have that
$$\int_{1}^e g(u) P(u)du = 0$$
for any polynomial $P(u)$. Now pick the polynomial to approximate $g$ to within an $\epsilon$ on $[1,e]$ and then take $\epsilon\to 0$ to obtain
$$\int_{1}^e g^2(u)du = 0$$
and it follows that $\frac{f(\log(u))}{u} = 0 \implies f\equiv 0$.