First some notation
$$\nabla \times \vec B \rightarrow \epsilon_{ijk}\nabla_j B_k$$
$$\nabla \cdot \vec B \rightarrow \nabla_i B_i$$
$$\nabla B \rightarrow \nabla_i B$$
Now, to your problem,
$$\nabla \cdot(\nabla \times \vec V)$$
writing it in index notation
$$\nabla_i (\epsilon_{ijk}\nabla_j V_k)$$
Now, simply compute it, (remember the Levi-Civita is a constant)
$$\epsilon_{ijk} \nabla_i \nabla_j V_k$$
Here we have an interesting thing, the Levi-Civita is completely anti-symmetric on i and j and have another term $\nabla_i \nabla_j$ which is completely symmetric: it turns out to be zero.
$$\epsilon_{ijk} \nabla_i \nabla_j V_k = 0$$
Lets make the last step more clear. We can always say that $a = \frac{a+a}{2}$, so we have
$$\epsilon_{ijk} \nabla_i \nabla_j V_k = \frac{1}{2} \left[ \epsilon_{ijk} \nabla_i \nabla_j V_k + \epsilon_{ijk} \nabla_i \nabla_j V_k \right]$$
Now lets interchange in the second Levi-Civita the index $\epsilon_{ijk} = - \epsilon_{jik}$, so that
$$\epsilon_{ijk} \nabla_i \nabla_j V_k = \frac{1}{2} \left[ \epsilon_{ijk} \nabla_i \nabla_j V_k - \epsilon_{jik} \nabla_i \nabla_j V_k \right]$$
Now we can just rename the index $\epsilon_{jik} \nabla_i \nabla_j V_k = \epsilon_{ijk} \nabla_j \nabla_i V_k$ (no interchange was done here, just renamed).
$$\epsilon_{ijk} \nabla_i \nabla_j V_k = \frac{1}{2} \left[ \epsilon_{ijk} \nabla_i \nabla_j V_k - \epsilon_{ijk} \nabla_j \nabla_i V_k \right]$$
We can than put the Levi-Civita at evidency,
$$\epsilon_{ijk} \nabla_i \nabla_j V_k = \frac{\epsilon_{ijk}}{2} \left[ \nabla_i \nabla_j V_k - \nabla_j \nabla_i V_k \right]$$
And, because V_k is a good field, there must be no problem to interchange the derivatives $\nabla_j \nabla_i V_k = \nabla_i \nabla_j V_k$
$$\epsilon_{ijk} \nabla_i \nabla_j V_k = \frac{\epsilon_{ijk}}{2} \left[ \nabla_i \nabla_j V_k - \nabla_i \nabla_j V_k \right]$$
And, as you can see, what is between the parentheses is simply zero.
