Differential Geometry-Hodge Star The Hodge star is given by $$*(dx^{i_1}\wedge dx^{i_2}\wedge....\wedge dx_{i_p})=\frac{1}{(n-p)!}e_{i_1 i_2....i_p i_{p+1}...i_n}dx^{i_{p+1}}\wedge dx^{i_{p+2}}\wedge....\wedge dx^{i_n}$$
The question is to prove $$**\omega_p=(-1)^{p(n-p)}\omega_p$$
by repeating the * operator on a p-form $\omega_p$. Now, how shall I prove it?
Here, $e_{ijk....}$ is antisymmetric tensor in n-dimension
 A: First of all, note that in the formula you give us,
$$*(dx^{i_1}\wedge dx^{i_2}\wedge\dots\wedge dx^{i_p})=\frac{1}{(n-p)!}e_{i_1 i_2\dots i_p i_{p+1}...i_n}dx^{i_{p+1}}\wedge dx^{i_{p+2}}\wedge....\wedge dx^{i_n},
$$
there is a summation involved! In fact, there are $n-p$ sums, since the indices are being contracted. I won't use Einstein's convention in this post. 
Explicitly we have:
$$*(dx^{i_1}\wedge dx^{i_2}\wedge\dots\wedge dx^{i_p})=\frac{1}{(n-p)!}\sum_{j_{p+1},\dots j_{n}=1}^ne_{i_1 i_2\dots i_p j_{p+1}...j_n}dx^{j_{p+1}}\wedge dx^{j_{p+2}}\wedge....\wedge dx^{j_n},
$$
where $e_{\cdots}$ is the Levi-Civita tensor. 
In fact you're summing the same $(n-p)$-form $(n-p)!$ times! This follows from the definition of the Levi-Civita tensor and that $dx_i\wedge dx_j = -dx_j\wedge dx_i$. 
To find how many times you're summing the same term observe that whenever an index is repeated, the term vanishes. In this case, the indices in the sum can take only $n-p$ distinct values and must all be different. So we get $(n-p)!$ possible combinations.
From this we get that (no sum involved!):
$$
*(dx^{i_1}\wedge dx^{i_2}\wedge\dots\wedge dx^{i_p})=e_{i_1 \dots i_n}dx^{i_{p+1}}\wedge dx^{i_{p+2}}\wedge....\wedge dx^{i_n}.
$$
Now, applying again the star operator we get:
\begin{align}
**(dx^{i_1}\wedge dx^{i_2}\wedge\dots\wedge dx^{i_p})
&=e_{i_1 \dots i_n}*dx^{i_{p+1}}\wedge dx^{i_{p+2}}\wedge....\wedge dx^{i_n}\\
&=e_{i_1\dots i_p i_{p+1}\dots i_n} e_{i_{p+1}\dots i_{n} i_1\dots i_{p}} dx^{i_1}\wedge dx^{i_2}\wedge\dots\wedge dx^{i_p}.
\end{align}
To conclude, observe that
$$
e_{i_1\dots i_p i_{p+1}\dots i_n} e_{i_{p+1}\dots i_{n} i_1\dots i_{p}} = (-1)^{p(n-p)}.
$$
(To show this count how many permutations you must do on the second term to obtain the first.)
