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. 

Similar to @wltrup with notational changes, we have
$$\begin{align}
\nabla \times \nabla \times \vec A&=(\partial_i \hat x_i)\times (\partial_j \hat x_j)\times (\hat x_k A_k)\\\\
&=\hat x_i\times(\hat x_j\times \hat x_k)\partial_i\partial_j(A_k) \tag 1\\\\
&=\left(\delta_{ik}\hat x_j-\delta_{ij}\hat x_k\right)\partial_i\partial_j(A_k)\tag 2\\\\
&=\hat x_j\partial_j\partial_iA_i-\hat x_k\partial^2_i(A_k) \tag 3\\\\
&=(\hat x_j \partial_j)(\partial_i A_i)-\partial^2_i(\hat x_kA_k) \tag 4\\\\
&=\nabla \nabla \cdot \vec A-\nabla^2\vec A \tag5
\end{align}$$
In going from $(1)$ to $(2)$ we made use of the vector triple product.  Note that $\delta_{ij}$ is the Kronecker Delta with $\delta_{ij}=1$ for $i=j$ and $0$ otherwise.
In going from $(2)$ to $(3)$, we used the sifting property of the Kronecker Delta.
In going from $(3)$ to $(4)$, we rearranged terms.
In going from $(4)$ to $(5)$, we recognized the terms of the final result in terms of their tensor representations.
