Showing that $\nabla\times(\nabla\times\vec{A}) = \nabla(\nabla\cdot\vec{A})-\Delta\vec{A}$ I have faced difficulties while trying to prove that
$$\nabla\times(\nabla\times\vec{A}) = \nabla(\nabla\cdot\vec{A})-\Delta\vec{A}$$ 
I don't have any clue how can I start to work with it. Any hint will be helpful. 

I am coming to this party about three years too late, but the answers posted here, and some here as well: enter link description here, may leave readers with the false impression that the vector Laplacian acting on a vector field (appearing on the right-hand side) always takes the form
$$\begin{align}
\nabla^2 \vec{A} = \nabla^2(A_i)\,\hat{e}_i ~~~ {\text{(summation convention in force)}} \tag 1
\end{align}$$
in an arbitrary basis and coordinate system of $\mathbb{R}^3$, where $\nabla^2$ is the Laplacian operator in the coordinates associated with the basis. This is true only in a cartesian coordinate system and its orthonormal basis. For example, in the cylindrical coordinate system $\{\rho, \phi, z\}$, with orthogonal (but not orthonormal) basis $\{\hat{e}_\rho, \hat{e}_\phi. \hat{e}_z\}$, it might be supposed from these derivations that
$$\begin{align}
\nabla^2 \vec{A} = \nabla^2(A_\rho)\,\hat{e}_\rho + \nabla^2(A_\phi)\,\hat{e}_\phi + \nabla^2(A_z)\,\hat{e}_z,
\end{align}$$
where
$$\begin{align}
\nabla^2(f) = \frac{1}{\rho}\frac{\partial}{\partial \rho}\left(\rho\frac{\partial f}{\partial \rho}\right) + \frac{1}{\rho^2}\frac{\partial^2 f}{\partial \phi^2} + \frac{\partial^2 f}{\partial z^2}
\end{align}$$
is the Laplacian operator in cylindrical coordinates acting on a scalar function $f$. This is quite false, as I first learned long ago from the textbook ``Mechanics of Deformable Bodies'', by the master mathematical physicist Arnold Sommerfeld. In the answers to Problems I.3 and I.4 of the book, he lays bare the fallacy of this assumption. The first two components of $\nabla^2{\vec{A}}$ contain extra terms besides $\nabla^2(A_\rho)$ and $\nabla^2(A_\phi)$.
In fact, Sommerfeld clearly states in Chapter I, page 23, what seems to have been overlooked, or at any rate, not emphasized, in many of the answers posted here: the formula $\nabla \times (\nabla \times \vec{A}) = \nabla(\nabla\cdot \vec{A}) - \nabla^2\vec{A}$ is an identity in the sense of (1) being true, only in cartesian coordinates. For any other coordinate system and basis, the formula serves to define $\nabla^2\vec{A}$ by
$$\begin{align}
\nabla^2\vec{A} := \nabla(\nabla\cdot \vec{A}) - \nabla \times (\nabla \times \vec{A}),
\end{align}$$
each term on the right-hand side being a well-defined operation in any coordinate system. So the upshot is that the formula can be shown to be true only in a cartesian coordinate system (as was done in the answers here); in any other coordinate system the formula instead defines the vector Laplacian acting on a vector field.
