How to prove that the spectrum of the Laplacian over $\Omega\subset \mathbb{R}^n$ is negative? I am looking for a proof of this well known fact and I guess it has to do with integration by parts (Green's identity). Unfortunately, I only know about 1-d integration by parts( I am just 3rd semester). could anybody sketch a short proof that shows that the spectrum of the Laplacian is negative?
My problem with the hint that brom gave me is that I do not know what I can assume about this surface integral and so on that appear in green's identity, but maybe I am looking at this from the wrong direction. 
If I start:
$$ \int_V u \Delta u + \underbrace{\langle \nabla u, \nabla u \rangle}_{\ge 0} = \int_{\partial V}u(\nabla u n) dS$$
I mean intuitively, the thing is, that if we assume that our function $\psi$ vanishes at infinity and we integrate over the whole space, then the surface integral is maybe zero. In this case, we had 
$$ \int_V u \Delta u=-\int_V\langle \nabla u, \nabla u \rangle \le 0$$
But I do not see the correct mathematical reasoning behind this!
 A: As brom already pointed out in the comments, for a bounded domain $\Omega\subset \mathbb{R}^n$, if $\lambda $ is a Dirichlet eigenvalue for Laplacian: 
$$
\Delta u = \lambda u \; \text{ in }\Omega,
\\
u = 0 \text{ on }\partial \Omega.\tag{1}
$$
Then
$$
\lambda \int_{\Omega} u^2 = \int_{\Omega} u\Delta u = - \int_{\Omega} |\nabla u|^2 + \color{blue}{\int_{\partial \Omega} u(\nabla u\cdot n)\,dS},
$$
where blue term vanishes and we get the Rayleigh quotient for Laplacian: for non-trivial $u$ solving (1)
$$
\lambda = - \frac{\int_{\Omega} |\nabla u|^2}{\int_{\Omega} u^2 } <0.
$$
For Neumann eigenvalues we have the first one being 0 and others being negative, in that there exists a non-trivial function $\Delta u =0 $ and $\nabla u\cdot n = 0$.


My problem with the hint that brom gave me is that I do not know what I can assume about this surface integral and so on that appear in Green's identity.

A key part here for either Dirichlet or Neumann eigenvalues for Laplacian, is to assume proper boundary conditions so that the boundary surface integral will vanish.
