Determinant of matrix over $\mathbb{F}_{p}[x]$ Let $p$ be a prime and  $C(x)$ be a following circulant matrix
$$
C(x) = \begin{pmatrix}  x^{p-1} & (x+1)^{p-1} & \cdots & (x+p-1)^{p-1} \\
(x+1)^{p-1} & (x+2)^{p-1} & \cdots & x^{p-1} \\
\vdots & \vdots & \ddots & \vdots \\
(x+p-1)^{p-1} & x^{p-1} & \cdots & (x+p-2)^{p-1}\end{pmatrix}
$$
where the polynomials are in $\mathbb{F}_{p}[x]$. I want to prove that the determinant of $C$ is $-1$, although I'm not sure if this is even true or not. What I know is that (i) $\det C(0) = -1$ and (ii) $\frac{d}{dx} \det C(x) = 0$. However, these two only shows that $\det C(x)$ has a form of $$\det C(x) = - 1 + \sum_{1\leq j \leq p-1} b_{j}x^{pj}$$
(note that degree of $\det C(x)$ is $\leq p(p-1)$), and I can't prove that the coefficients $b_{j}$'s are actually zero. Is it possible to proceed in this direction or are there any better way to prove $\det C(x) = -1$?
 A: Let us (for a moment) work instead with the standard circulant matrix
$$
D(x)=\sum_{k=0}^{p-1} (x+k)^{p-1} P^k,
$$
where
$$
P=\begin{pmatrix}
0 & 1 & 0 & \dots & 0\\
0 & 0 & 1 & \dots & 0\\
\vdots &\vdots &\vdots &\ddots&\vdots\\
0 & 0 & 0 & \dots & 1\\
1 & 0 & 0 & \dots & 0\\
\end{pmatrix}
$$
is the permutation matrix. Note that $\det P=1$.
By re-ordering the columns we see that $\det C(x)=(-1)^{\frac{p-1}{2}}\det D(x)$.
By considering $D(x)P$ we see that $\det D(x+1)=\det D(x)$. The same is then true for $C(x)$, and by repeating the argument we have that
$$\det C(x+k)=\det C(x)\ \  \text{   for  } k=0,1,\dots,p-1. \tag{*}$$
You have already proved that
$$
\det C(x) = -1+\phi(x^p)
$$
where $\phi(t)\in\mathbb{F}_p[t]$ is of degree at most $(p-1)$.
You also have
$$\phi(0)=0$$
and so by (*) we will get
$$\phi(k)=0\ \  \text{   for  } k=0,1,\dots,p-1. $$
We therefore have that $\phi(t)$ is a multiple of the polynomial
$$\prod_{k\in\mathbb{F}_p}(t-k)=t^p-t.$$
But as the degree of $\phi$ is at most $p-1$ this multiple must  be $0$.
Hence
$$\det C(x)=-1$$
as required.
