Let $\Gamma_p : \mathbb{Z}_p \to \mathbb{Z}_p^{\times}$ be the $p$-adic gamma function.  I thought that I had successfully calculated $\Gamma_p(1 - 1/4)$, but sage is telling me I'm wrong (this is more than a sign error, which I have probably made along the way).  Here is what I've done:

Let $p > 2$ be a prime such that $p \equiv 1 \bmod 4$, and $q = p^2$.  We will assume that all roots of unity lie in $\mathbb{C}_p^{\times}$.   Let $\omega_q : \mathbb{F}_q^{\times} \to \mu_{q-1}\subseteq \mathbb{Z}_p[\mu_q]$ be the Teichmüller character and $\zeta$ a primitive $p$-th root of unity.  We denote $\chi = \omega_q^{(q-1)/4}$.  Let $\operatorname{Tr} : \mathbb{F}_q \to \mathbb{F}_p$ and $N : \mathbb{F}_q \to \mathbb{F}_p$ be the trace and norm respectively.  For a finite field $F$ and multiplicative character $\psi$ on $F$, the Gauss sum for $\psi$ is defined to be
\begin{align*}
G(\psi) = \sum_{c \in F} \psi(c) \zeta^{\operatorname{Tr}_{F/\mathbb{F}_p}(c)}
\end{align*}
and the Jacobi sum for $\psi$ is defined
\begin{align*}
J(\psi) = \sum_{c \in F} \psi(c) \psi(1-c).
\end{align*}
 Let $\pi$ be the maximal prime in $\mathbb{Z}_p(\zeta)$ such that $\pi \equiv 1-\zeta \mod{\pi^2}$.  

It follows from the Gross–Koblitz formula that
\begin{align*}
\Gamma_p\left(1 - \frac{1}{4}\right) ^2= -\frac{\pi^{(p-1)/2}G(\chi)}{q}.
\end{align*}

Also, if $\omega = \omega_q \rvert_{\mathbb{F}_p} : \mathbb{F}_p \to \mu_{p-1}$ and $\psi = \omega^{(p-1)/4}$, then since
\begin{align*}
\chi(c) = \omega_q^{(p-1)/4}(c^{p+1})  = \omega_q^{(p-1)/4}(N(c)) = \omega^{(p-1)/4}(N(c))
\end{align*}
$G(\chi) = G^2(\psi)$ by the Hasse–Davenport lifting relation.

Now, $\psi$ is a quartic character on $\mathbb{F}_p$, which means $\psi^2$ a quadratic character.  So, from the well known value  
\begin{align*}
G(\psi^2) = \pi^{(p-1)/2}  
\end{align*}
and from the relation $J(\psi) = G^2(\psi)/G(\psi^2)$, we have
\begin{align*}
G^2(\psi) =  \pi^{(p-1)/2} J(\psi).
\end{align*}

Putting the pieces together, we now have
\begin{align*}
\Gamma_p\left(1 - \frac{1}{4}\right) ^2= \frac{J(\psi)}{p}.
\end{align*}

Again, this all seems correct to me, but sage is telling me I'm wrong.  I've been looking at this for a while, and I can't seem to figure out where the mistake is.  My first guess is that I messed up the Hasse–Davenport part, but I can't quite see how.