Help simplifying vector calculus

I'm working on a problem and I have come to the following: $$\nabla \cdot (\nabla \vec{u} + (\nabla\vec{u})^T)$$ $$ = \nabla \cdot \nabla\vec{u} + \nabla \cdot (\nabla\vec{u})^T$$ From knowing the solution of the problem, this is supposed to simplify to: $$\nabla^2\vec{u}$$ But I can't see how this is done. I'm pretty sure I did the first step right?

The $(i,j)$-component of $\nabla \vec{u} + (\nabla \vec{u})^T$ is $$\partial_j u_i + \partial_i u_j$$ So the $j$th entry of $\nabla\cdot (\nabla \vec{u} + (\nabla \vec{u})^T)$ is (in Einstein convention) $$\partial_i(\partial_j u_i + \partial_i u_j) = \partial_j(\partial_i u_i) + \partial_i\partial_i u_j$$ The latter is the $j$th entry of $\nabla(\nabla\cdot \vec{u}) + \nabla^2\vec{u}$, so $$\nabla\cdot (\nabla\vec{u} + (\nabla \vec{u})^T) = \nabla(\nabla\cdot \vec{u}) + \nabla^2\vec{u}$$ You need to have $\nabla \cdot \vec{u} = 0$ in order to simplify to $\nabla^2\vec{u}$. This will hold true, e.g., when $\vec{u}$ is the fluid velocity field of an incompressible flow.