How to find the Newton polygon of the polynomial product $ \ \prod_{i=1}^{p^2} (1-iX)$ How to find the Newton polygon of the polynomial product   $ \ \prod_{i=1}^{p^2} (1-iX)$ ?
Answer:
Let $ \ f(X)=\prod_{i=1}^{p^2} (1-iX)=(1-X)(1-2X) \cdots (1-pX) \cdots (1-p^2X).$
If I multiply , then we will get a polynomial of degree $p^2$.
But it is complicated to express it as a polynomial form.
So it is complicated to calculate the vertices $ (0, ord_p(a_0)), \ (1, ord_p(a_1)), \ (2, ord_p(a_2)), \ \cdots \cdots$ 
of the above product.
Help me doing this
 A: It’s really quite simple. There are $p^2-p$ roots $\rho$ with $v(\rho)=0$, $p-1$ roots with $v(\rho)=-1$, and one root with $v(\rho)=-2$. Consequently, there is one segment of the polygon with slope $0$ and width $p^2-p$, one segment with slope $1$ and width $p-1$, and one segment with slope $2$ and width $1$.
Thus, the vertices are $(0,0)$, $(p^2-p,0)$, $(p^2-1,p-1)$, and $(p^2,p+1)$.
A: Partial Answer: regarding the coefficients of the polynomial: 
Fix one term in the brackets, say $Y=(1-5X)$. In order for the coefficient $5$ to contribute to $a_j$, we have to multiply $Y$ with $j-1$ other brackets, since this is the only way of getting a power of $j$ for $X$. This corresponds to choosing a subset $S \in \{1,2,\ldots,p^{2}\}$ of size $j-1$ since each term in the product has a unique coefficient for $X$ that is in $\{1,2,\ldots,p^{2}\}$. This leads to 
\begin{equation}
a_j=(-1)^{j} \underset{  S \subset \{1,2, \ldots, p^{2} \}, \ |S|=j}{\sum} \prod \limits_{s \in S} s \ . 
\end{equation}
