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. 

All you need to know is the definitions of (just taking $\mathbb{R}^3$ for simplicity)
$$
\nabla=\left[\matrix{\frac{\partial}{\partial x}\\ \frac{\partial}{\partial y}\\ \frac{\partial}{\partial z}}\right]\quad\text{(a symbolic vector)},\qquad
\vec A=\left[\matrix{A_1(x,y,z)\\A_2(x,y,z)\\A_3(x,y,z)}\right]\qquad\text{(a vector)}
$$
the scalar product and the vector product of two vectors. Well, also $\Delta=\nabla\cdot\nabla$. Then just apply the products formally to the vector coordinates and keep the order so that the derivatives affect functions.
