p-adic gamma function

Remarks

The positive integers are dense in $\Z_p$, so such a continuous extension is unique. It exists because $\Gamma_p(m)$ and $\Gamma_p(n)$ are $p$-adically close when $m$ and $n$ are; the proof uses a generalisation of Wilson's theorem. The values of $\Gamma_p$ are units of $\Z_p$.

Example: for $p = 3$, $\Gamma_3(1) = -1$, $\Gamma_3(2) = 1$, $\Gamma_3(3) = -2$, $\Gamma_3(4) = 2$ and $\Gamma_3(5) = -8$. Each value follows from the previous one by the rule $$\Gamma_p(x + 1) = \begin{cases} -x\, \Gamma_p(x) & \text{if } x \in \Z_p^\times, \\ -\Gamma_p(x) & \text{if } x \in p\Z_p, \end{cases}$$ which holds for all $x \in \Z_p$; here $\Gamma_p(0) = 1$.

Non-example: the factorial $n \mapsto n!$ has no continuous extension to $\Z_p$. Indeed $1 + p^k \to 1$ in $\Z_p$ as $k \to \infty$, but $(1 + p^k)! = (1 + p^k) \cdot (p^k)!$ is divisible by $p^k$ and so tends to $0$, while $1! = 1$.