Codifferential of a $p$-vector in components I'm learning differential geometry from a textbook, and I got stuck on a problem. 
I'm supposed to calculate this for a $p$-vector $F$ in $n$ dimensions:
$(\mathrm{div}_\omega F)^{i...j} = F^{ki...j}\ _{,k}$
where
$\mathrm{div}_\omega F = (-1)^{n(p-1)} * \mathrm{d} * F$
The star is the Hodge star; the RHS operator is the codifferential. 
I got as far as 
$\mathrm{d}*F = \frac{1}{p!(n-p)!} \varepsilon_{i...jk...l}F^{i...j}\ _{,m} \ \mathrm{d}x^m\wedge\mathrm{d}x^k\wedge...\wedge\mathrm{d}x^l$
Clearly, $m \in {i...j}$ for all non-zero terms. But I can't get any further than that, and not for lack of trying. Any tips will be appreciated. :)
 A: Let us denote the $p$-vector $F$ in local coordinates by $$F=F^{i_1\cdots i_p}(x)e_{i_1}\wedge\cdots\wedge e_{i_p} $$
and let us suppose to work on a flat manifold of dimension $n$: the metric tensor $g$ is just the identity matrix in $n$ dimensions.  We will always use the Einstein notation for repeated indices. Using the above conventions the $n-p$-vector $\star F$ becomes
$$\star F=\frac{1}{(n-p)!}\epsilon_{i_1\cdots i_n}F^{i_1\cdots i_p}(x)e_{i_{p+1}}\wedge\cdots\wedge e_{i_n} $$
and so
$$d\star F=\frac{1}{(n-p)!}\sum_{j=1}^p\epsilon_{i_1\cdots i_n}\frac{\partial F^{i_1\cdots i_p}(x)}{\partial x_{i_j}}e_{i_j}\wedge e_{i_{p+1}}\wedge\cdots\wedge e_{i_n}. $$
In the above formula the sum is up to $p$ as $ e_{i_j}\wedge e_{i_{p+1}}\wedge\cdots\wedge e_{i_n}=0$ if $j\in\{p+1,\dots, n\}$ because $e_\bullet\wedge e_\bullet =0$
by antisymmetry of the wedge product. We end up with the $n-(n-p+1)=p-1$ vector
$$\star d\star F=\frac{1}{(p-1)!(n-p)!}\sum_{j=1}^p\epsilon_{i_1\cdots i_n}\frac{\partial F^{i_1\cdots i_p}(x)}{\partial x_{i_j}}\star\left(e_{i_j}\wedge e_{i_{p+1}}\wedge\cdots\wedge e_{i_n}\right)= \\
\frac{1}{(p-1)!(n-p)!}\sum_{j=1}^p\epsilon_{i_1\cdots i_n}\epsilon_{i_j i_{p+1}\dots i_n i_1\cdots \hat{i}_j\cdots i_p }\frac{\partial F^{i_1\cdots i_p}(x)}{\partial x_{i_j}}e_{i_{1}}\wedge\cdots\wedge \hat{e}_{i_j}\wedge\cdots\wedge e_{i_p} ~~(*)
 $$
where $\hat{\cdot}$ denotes omission. I won't manipulate that formula further but I will provide you with an easier example. 
Let us consider the $n=3$ and $p=1$ case, with $$F=F^{i_1}(x)e_{i_1}=F^1(x)e_1+F^2(x)e_2+F^3(x)e_3.$$ Then (*) becomes $(j=1=p)$
$$\star d\star F=\frac{1}{2!}\epsilon_{i_1 i_2 i_3}\epsilon_{i_1 i_2 i_3}\frac{\partial F^{i_1}(x)}{\partial x_{i_1}}=
\frac{1}{2}\left[(\epsilon_{1 23}\epsilon_{123}+\epsilon_{1 32}\epsilon_{132})\frac{\partial F^{1}(x)}{\partial x_{1}} +
(\epsilon_{2 13}\epsilon_{213}+\epsilon_{2 31}\epsilon_{231})\frac{\partial F^{2}(x)}{\partial x_{2}} + \right. \\
\left.(\epsilon_{3 21}\epsilon_{321}+
\epsilon_{3 12}\epsilon_{312})\frac{\partial F^{3}(x)}{\partial x_{3}}
\right]=\frac{\partial F^{1}(x)}{\partial x_{1}}+ \frac{\partial F^{2}(x)}{\partial x_{2}} + \frac{\partial F^{3}(x)}{\partial x_{3}},
 $$
i.e. the divergence of the vector field $F$.
