I'm trying to show that $\sum_{i = 0}^{p-2} (i+1)^{-1} t^{i+n}$ where $0 \leq n \leq p-2$ spans the vector space $\mathbb{F}_p[t]/(1-t)^{p-1}$ as a rank $p-1$ module over $\mathbb{F}_p$. 

In other words, I would like to show that the determinant of the following matrix is a unit in $\mathbb{F}_p$. I've shown this for $p = 2, 3, 5, 7, 11, 13$. I have tried to use induction but failed.  This is such a natural matrix I am hoping someone recognizes it! 

$$\begin{pmatrix} 1 & 0 & -1 & (p-2)^{-1} & \cdots  & 4^{-1} & 3^{-1} \\
2^{-1} &  1 & 0 & -1 & (p-2)^{-1} & \cdots & 4^{-1} \\
3^{-1} & 2^{-1} & 1 & 0 & -1 & \ddots & \vdots  \\
\vdots & 3^{-1} & 2^{-1} & 1 & 0  & \ddots & (p-2)^{-1} \\
 (p-3)^{-1} & \ddots & \ddots & \ddots & \ddots & 0  & -1 \\
(p-2)^{-1} & (p-3)^{-1} & (p-4)^{-1} &  \ddots & 2^{-1} & 1 & 0 \\
-1 & (p-2)^{-1} & (p-3)^{-1} & \cdots & 3^{-1} & 2^{-1} & 1
\end{pmatrix}$$


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Edit: At risk of overcrowding the above question, I am adding some context below the line. Feel free to ignore it.

Let us look at the action of $G := C_p \rtimes C_{p-1} \simeq (\mathbb{F}_p, +) \rtimes (\mathbb{F}_p^*, \times)$. Then, $G$ acts on $x \in X := F_p$ as follows, $(c, m)(x) = c + mx$. Let R be a $\mathbb{Z_p}$-algebra, and $A=R[C_p]$ be the permutation representation. If $x \in X$, we write $[x]$ as the corresponding element in the module. Let us fix $\sigma := (1,1)$ and $\tau := (0, a)$ to be the generators of $G$, where $a$ is a fixed primitive root of $\mathbb{Z}/p$.

Now, let $B$ be the kernel of the augmentation map $R[X] \to R$. This is a free representation of rank $p-1$ over $R$. Further, as a $C_p$-module, $B$ is isomorphic to $R[X]/N$, where $N$ is generated by $(1+\sigma+\cdots+\sigma^{p-1})$. Further, $N$ is isomorphic to the trivial representation as a $C_p$ module. 

Let $V$ be a rank 1 $C_{p-1}$-subrepresentation of $B$. We may extend the inclusion map $V \to B$ of $C_{p-1}$ representations to a map of $G$ representations: $$f: R[G] \otimes_{R[C_{p-1}]} V \to B.$$  

I wish to pick $V$ such that this map is surjective. By Nakayama's lemma, since we are working with local rings, it suffices to show this map is surjective mod $p$. 

To get to the below phrasing, I rewrote $R[X]$ as $R[t]/(t^p-1)$. Here, $\sigma$ acts by taking $t^{i} \mapsto t^{i+1}$. Then, if we consider $B$ mod $p$, which we call $\overline{B}$, then $\overline{B} = \mathbb{F}_p[t]/(t-1)^{p-1}$. There are $(p-1)$ 1-d $C_{p-1}$ subrepresentations of $B$, where $t$ in $\mathbb{F}_p^\times$ acts by multiplication with $t^r$, where $1 \leq r \leq p-1$. Further, from trial and error I noticed that picking $V$ to be the subrepresentation where $r = 1$, seems to make $f$ surjective. In other words, choosing $V$'s generator mod p to be $y := \sum_{j = 0}^{p-2}a^{-j}t^{a^k} = \sum_{i=0}^{p-2} i^{-1}t^i$ seems to make $f$ surjective. Note that the image of $f$ is $(y, \sigma(y), \cdots, \sigma^{p-2}(y))$.