Would anybody mind teaching me how to work these indices?
Definitions:
Throughout the following, repeated indices are to be summed over.
Hodge dual of a p-form $X$: $$(*X)_{a_1...a_{n-p}}\equiv \frac{1}{p!}\epsilon_{a_1...a_{n-p}b_1...b_p}X^{b_1...b_p}$$ Exterior derivative of p-form $X$: $$(dX)_{a_1...a_{p+1}}\equiv (p+1) \nabla_{[a_1}X_{a_2...a_{p+1}]}$$
Given the relation $$\epsilon^{a_1...a_p c_{p+1}...c_n}\epsilon_{b_1...b_pc_{p+1}...c_n}\equiv p!(n-p)! \delta^{a_1}_{[b_1}...\delta^{a_p}_{b_p]}\,\,\,\,\,\,\,\,\,(\dagger)$$ where $\epsilon_{a_1...a_n}$ is an orientation of the manifold.
Why then is $$(*d*X)_{a_1...a_{p-1}}=(-1)^{p(n-p)}\nabla^b X_{a_1...a_{p-1}b}$$?
How far I've got myself:
Firstly, I believe $(*d*X)$ means $*(d(*X))$? $$(d*X)_{c_1...c_{n-p+1}}=\frac{n-p+1}{p!}\nabla_{[c_1}\epsilon_{c_2...c_{n-p+1}]b_1...b_p}X^{b_1...b_p}$$ Then $$*(d*X)_{d_1...d_{p-1}}=\frac{n-p+1}{(n-p+1)!p!}\epsilon_{d_1...d_{p-1}c_1...c_{n-p+1}}\nabla^{[c_1}\epsilon^{c_2...c_{n-p+1}]b_1...b_p}X_{b_1...b_p}$$ $$=\frac{1}{(n-p)!p!}\epsilon_{d_1...d_{p-1}c_1...c_{n-p+1}}\nabla^{[c_1}\epsilon^{c_2...c_{n-p+1}]b_1...b_p}X_{b_1...b_p}$$
Now I know that I should apply $(\dagger)$ but I don't know how to given the antisymmetrisation brackets. Would someone mind explaining it to me please? Thank you!