Let $V$ be a complete discrete valuation ring whose residue field is a finite field $k=\mathbf{F}\_q$. Let $\pi\in V$ be a uniformizer.


For any integer $d,n\ge 0$, define:


$${\pi^d \choose n} := \frac{\pi^d\cdot(\pi^d -1)\cdot\ldots\cdot(\pi^d-n+1)}{n!}.$$



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> * Is ${\pi^d\choose n}$ an element of $V$?
> * For exactly what integers $n\ge 0$ is ${\pi^d\choose n}$ a unit?
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**Example.** If $V = \mathbf{Z}\_p$, $\pi = p$, then ${p^d\choose n}$ is zero unless $0\le n\le p^d$, and its $p$-adic valuation is $d-v\_p(n)\ge 0$ for $1\le n\le p^d$. In all cases, the answer to the first question is yes, and the answer to the second question is: for $n = 0, c\cdot p^d$, with $(c,p) = 1$.



Binomial coefficients in discrete valuation rings
If $p=2=n$, $d=1$, $V=\mathbf{Z}\_2[\sqrt{2}]$ and $\pi=\sqrt{2}$, then
$$ \binom{\pi^d}{n} = \frac{\sqrt{2}(\sqrt{2} - 1)}{2} = 2^{-1/2}\cdot \mathrm{unit} \notin V. $$


