Hi,

An interesting question! Not an answer, but the necessary methods almost certainly are known and exist somewhere; maybe this will help: the Laplace transform gives (a multiple of) a unitary operator from $L^2(0, \infty)$ onto the Hardy space $H^2( \{ \mathrm{Re}(z)>0 \})$ (depending on your normalisations).

The Laplace transform of $x^{n} e^{-x/n}$ is, up to a constant, $(z+1/n)^{-(n+1)}$. 

So you're asking whether the orthogonal complement of these functions is zero in $H^2$. The scalar product of $F(z)$ with $(z+\overline{\lambda})^{-k-1}$ is, up to a constant, the derivative $F^{(k)}(\lambda)$.

Thus (assuming my algebra is correct), your question is equivalent to asking whether there is any non--trivial $F \in H^2$ satisfying

$$
F^{(n)}(1/n) = 0, \qquad n=1,2,3,\ldots
$$

This is a question in Complex Analysis. If you just wanted some analytic function $F$ on the half-place $ \{ x+iy : x>0 \}$  to satisfy this, it's possible (we can prescribe finitely many derivatives at all points of any countable set without limit point in the domain). The extra condition $F \in H^2$ is the difficult part.

Similar-looking questions have well-known answers. For example: there is no non-trivial $G \in H^2$ satisfying $G(z_n)=0$ if and only if $\sum_n \frac{\mathrm{Re}(z_n)}{|1+z_n|^2} = +\infty$ (the Blaschke condition). e.g. consider $z_n$ going to zero, or out to infinity; if it does so quickly that the sum is finite, then the set $\{ z_n \}$ is sparse enough to allow non-trivial $G$.

I'll check again later - must rush off now to a tedious meeting!!