I'm reading from Albert Schwartz "Topology for Physicists". Everything is on a compact, oriented, n-dimensional Riemannian manifold. He defines the dual of an alternating rank $k$ tensor $F_{i_1...i_k}$ as $\bar{F}_{j_1...j_{n-k}} = \frac{1}{k!}\sqrt{g}\epsilon_{j_1...j_{n-k}i_1...i_k}F^{i_1...i_k}.$ $g$ is the determinant of the metric, and $\epsilon$ is the Levi-Civita symbol. Then he defines the dual of a $k$-form $\omega = \frac{1}{k!}F_{i_1...i_k}dx^{i_1}\wedge...\wedge dx^{i_k}$ as $*\omega = \frac{1}{(n-k)!}\bar{F}_{j_1...j_{n-k}}dx^{j_1}\wedge...\wedge dx^{j_{n-k}}.$ Then he says that $**\omega = (-1)^{k(n-k)}\omega$. I am trying to prove this, but my most simplification does not get rid of the $g$ term.
I use $**\omega = *(\frac{1}{(n-k)!}\bar{F}_{j_1...j_{n-k}}dx^{j_1}\wedge...\wedge dx^{j_{n-k}}) = \frac{1}{k!}\bar{\bar{F}}_{i_1...i_k}dx^{i_1}\wedge...\wedge dx^{i_k}$, so I just need to compute $\bar{\bar{F}}$.
$\bar{\bar{F}}_{i_1...i_k} = \frac{1}{(n-k)!}\sqrt{g}\epsilon_{i_1...i_kj_1...j_{n-k}}\bar{F}^{j_1...j_{n-k}} = \frac{1}{(n-k)!}\sqrt{g}\epsilon_{i_1...i_kj_1...j_{n-k}}g^{j_1r_1}...g^{j_{n-k}r_{n-k}}\bar{F}_{r_1...r_{n-k}} = \frac{1}{(n-k)!}\sqrt{g}\epsilon_{i_1...i_kj_1...j_{n-k}}g^{j_1r_1}...g^{j_{n-k}r_{n-k}}\frac{1}{k!}\sqrt{g}\epsilon_{r_1...r_{n-k}q_1...q_k}F^{q_1...q_k} = \frac{1}{(n-k)!}\sqrt{g}\epsilon_{i_1...i_kj_1...j_{n-k}}g^{j_1r_1}...g^{j_{n-k}r_{n-k}}\frac{1}{k!}\sqrt{g}\epsilon_{r_1...r_{n-k}q_1...q_k}g^{q_1p_1}...g^{q_kp_k}F_{p_1...p_k}.$
At this point I try to shuffle around the metric terms and raise the indices of the Levi-Civita symbols, but I'm not sure how to get everything to cancel out nicely. I've seen an alternate proof of this by showing that $\omega \wedge *\omega$ is the volume form for any $\omega$, but I get equally stuck in this direction. I'd prefer answers that proceed in this computation, rather than hand-waving or using some other definition of the dual form, but any help is appreciated.
$\newcommand{\sgn}{\mathrm{sgn}}$
You will need the generalized Kronecker delta. Also, remember you can't simply raise or lower the indices of the Levi-Civita symbols, because they are not tensors. To remedy this, use the Levi-Civita tensors instead
$$E_{i_1\dots i_n} = \sqrt{g} \,\epsilon_{i_1\dots i_n}$$ $$E^{i_1\dots i_n} = \frac{\sgn(g)}{\sqrt{g}}\epsilon^{i_1\dots i_n}$$ where $\sgn(g)$ is the sign of the metric, i.e.: the product of its signature. Hence we have $$\frac{1}{k!(n-k)!}\sqrt{g}\epsilon_{i_1\dots i_kj_1\dots j_{n-k}}\sqrt{g}\epsilon_{r_1\dots r_{n-k}q_1\dots q_{k}}g^{j_1r_1}\cdots g^{j_{n-k}r_{n-k}} g^{q_1p_1}\cdots g^{q_kp_k}F_{p_1\dots p_k} \\=\frac{1}{k!(n-k)!}E_{i_1\dots i_kj_1\dots j_{n-k}}E_{r_1\dots r_{n-k}q_1\dots q_{k}}g^{j_1r_1}\cdots g^{j_{n-k}r_{n-k}} g^{q_1p_1}\cdots g^{q_kp_k}F_{p_1\dots p_k}$$ and then we can raise and lower indices freely. $$=\frac{1}{k!(n-k)!}E_{i_1\dots i_kj_1\dots j_{n-k}}E^{r_1\dots r_{n-k}q_1\dots q_{k}}\delta^{j_1}_{r_1}\cdots \delta^{j_{n-k}}_{r_{n-k}} \delta_{q_1}^{p_1}\cdots \delta_{q_k}^{p_k}F_{p_1\dots p_k}\\ =\frac{1}{k!(n-k)!}E_{i_1\dots i_kr_1\dots r_{n-k}}E^{r_1\dots r_{n-k}p_1\dots p_{k}}F_{p_1\dots p_k}\\ =\frac{\sgn(g)}{k!(n-k)!}\delta_{i_1\dots i_kr_1\dots r_{n-k}}^{r_1\dots r_{n-k}p_1\dots p_{k}}F_{p_1\dots p_k}\\ =\frac{\sgn(g)}{k!(n-k)!}(-1)^{k(n-k)}\delta_{i_1\dots i_kr_1\dots r_{n-k}}^{p_1\dots p_{k}r_1\dots r_{n-k}}F_{p_1\dots p_k}\\ =\frac{\sgn(g)}{k!}(-1)^{k(n-k)}\delta_{i_1\dots i_k}^{p_1\dots p_{k}}F_{p_1\dots p_k}\\ =\sgn(g)(-1)^{k(n-k)}F_{i_1\dots i_k}$$ which reduces to your formula in the case $\sgn(g)=1$.
The crucial steps are $$E_{i_1\dots i_kr_1\dots r_{n-k}}E^{r_1\dots r_{n-k}p_1\dots p_{k}} = \sgn(g)\,\delta_{i_1\dots i_kr_1\dots r_{n-k}}^{r_1\dots r_{n-k}p_1\dots p_{k}}$$ $$\delta_{i_1\dots i_kr_1\dots r_{n-k}}^{r_1\dots r_{n-k}p_1\dots p_{k}} = (-1)^{k(n-k)}\delta_{i_1\dots i_kr_1\dots r_{n-k}}^{p_1\dots p_{k}r_1\dots r_{n-k}}$$ $$\delta_{i_1\dots i_kr_1\dots r_{n-k}}^{p_1\dots p_{k}r_1\dots r_{n-k}} = (n-k)!\,\delta_{i_1\dots i_k}^{p_1\dots p_{k}}$$ and $$\delta_{i_1\dots i_k}^{p_1\dots p_{k}}F_{p_1\dots p_k} = k!\,F_{[i_1\dots i_k]}$$ which follow from the wiki article I cited above.
On a side note: this kind index-heavy manipulations which only involve tensors are more bearable in Penrose's graphical notation, in case you want to take a look.