Determinant of prime cyclotomic field. In this paper, Example 5.3, I would like to know how he simplified the Vandermonde determinant. What I am asking is how did he get the steps. I feel like there are some steps missing. Considering there are 7 steps in the solution (7 '='):
Q1. Firstly how did he split the product in to two products, what is the theory? (Step 2)
Q2.  How did one of the products disappear? What did he apply? (Step 3)
Q3. In Step 5, how does the Step 4 term (with products in the numerator and denominator) evaluate to $-1/p$
Q4. Also the reasoning since $p-1$ is odd doesn't add up as $p-2$ term should remain and not $p-1$ as that is even?
Please kindly explain this proof. Thank you.
 A: There are a number of typos in the source, but here goes.
(1) In more detail:
$$\begin{array}{ll}
\displaystyle \prod_{i<j}(a_i-a_j)^2 & \displaystyle =\prod_{i<j}(a_i-a_j)(a_i-a_j) \\ & =\displaystyle \prod_{i<j}(-1)(a_j-a_i)(a_i-a_j) \\ & =(-1)^N\displaystyle\prod_{i<j}(a_j-a_i)(a_i-a_j) \\ & = \displaystyle(-1)^N\left[\prod_{i<j} (a_j-a_i)\right] \left[\prod_{i<j}(a_i-a_j)\right] \\ & = \displaystyle(-1)^N \left[\prod_{j<i} (a_i-a_j)\right] \left[\prod_{i<j}(a_i-a_j)\right] \\ & = \displaystyle (-1)^N \prod_{i\ne j}(a_i-a_j) \\ & = \displaystyle (-1)^N \prod_i\left[\prod_{\substack{i\ne j \\ i\textrm{ fixed}}} (a_i-a_j)\right]
\end{array} $$
This is not two products, it's a product of products. Letting $(i,j)$ range over all possible values with $i\ne j$ is the same as letting $i$ range over all values, and within each possible value of $i$, letting $j$ range over every value except for $i$.
(Above $N$ stands for the number of pairs $(i,j)$ with $i<j$, which is $p-1$ choose $2$.)
(2) We know $X^p-1=(X-1)\prod_j(X-\zeta_j)$. Taking the derivative with the product rule and evaluating at $X=\zeta_i$ yields the formula $-p\zeta_i^{p-1}/(1-\zeta_i)=\prod_{j\ne i}(\zeta_i-\zeta_j)$ (with $i$ fixed).
(3) I assume you know where $(-p)^{p-1}$ comes from. In the numerator, 
$$\prod_i \zeta_i^{p-1}=\left(\prod_i \zeta_i\right)^{p-1}=(\zeta^{1+\cdots+(p-1)})^{p-1}=(\zeta^p)^{(p-1)^2/2}=1.$$
In the denominator,
$$\prod_i (1-\zeta_i)=\left.\frac{1-X^p}{1-X}\right|_{X=1}=1^{p-1}+\cdots+1^0=p. $$
(4) In the exponent of $(-1)$, arithmetic is mod $2$. If $p$ is odd, then $p-2\equiv 1$ mod $2$, and you need to keep $p-1$ in order for $(p-1)/2$ to make sense mod $2$.
