Green's First Identity involving Electric Potential

I am attempting to work through a paper that involves some slightly unfamiliar vector calculus, as well as many omitted steps. It begins with the potential energy due to an electric field, familiarly expressed as:
[tex]
U_{el} = \frac{\epsilon_r\epsilon_0}{2} \iiint_VE^2dV = \frac{\epsilon_r\epsilon_0}{2} \iiint_V(\nabla{\Phi})^2dV
[/tex]
I am not quite sure as to the expression for a squared gradient, but I worked from using ##\nabla(\Phi\nabla\Phi) = (\nabla\Phi)^2 + \Phi\nabla^2\Phi## so that ##(\nabla\Phi)^2 = \nabla(\Phi\nabla\Phi) - \Phi\nabla^2\Phi##. Now, I can rewrite ##U_{el}## as:
[tex]
U_{el} = \frac{\epsilon_r\epsilon_0}{2}\Big[\iiint_V\nabla(\Phi\nabla\Phi)dV - \iiint_V\Phi\nabla^2\Phi{dV}\Big]
[/tex]
I recognize the presence of Poisson's equation in the second term, but as for the first? This is where I am stuck. I am supposed to use these expressions together with the Divergence theorem so that I can invoke Green's first identity:
[tex]
\int_{\Omega}(\psi\nabla^2\phi + \nabla\phi\cdot\nabla\psi)dV = \oint_{\Omega}\psi(\nabla\phi\cdot{\bf{n}})dS
[/tex]
For two scalar functions ##\psi## and ##\phi##. Ultimately, I need to arrive at the expression:
[tex]
U_{el} = \frac{1}{2}\oint_S\sigma\phi_0dS + \frac{1}{2}\iiint_V\rho\Phi{dV}
[/tex]
Using Poisson's equation ##\nabla^2\Phi = -\frac{\rho}{\epsilon_0}## and ##\sigma = -{\epsilon_r\epsilon_0}\frac{\partial{\Phi}}{\partial{n}}## and using the fact that ##\phi_0## is the potential at the surface.

But I do not have any dot products associated with what I did above and so I do not see how to proceed; the only way I see how this could work is understanding that there is really only one scalar function here instead of two, namely ##\Phi##, but I don't see this allowing me to make ##\nabla(\Phi\nabla\Phi) = \nabla\Phi\cdot\nabla\Phi##. Am I forgetting something simple?