A conjecture on binomial coefficients and roots of unity Is the following true?
Let $p$ be a prime and let $w$ be a $(p-1)$st root of unity (not necessarily primitive). Then
$$\binom{w}{n}=\frac{w(w-1)\cdots(w-n+1)}{n!}$$ is $p$-integral; i.e., it can be expressed as a polynomial in $w$ with $p$-integral coefficients. In other words, if $\Phi_k(x)$ is the $k$th cyclotomic polynomial and $k$ divides $p-1$ then the remainder of $\binom{x}{n}$ modulo $\Phi_k(x)$ is $p$-integral.
 A: Here is an elementary and explicit way to see this:
Suppose we have a set of $p$ integers $A=\{a_1, a_2,\dots, a_p\}$ which forms a complete set of residues modulo p. Then we have
$$\prod_{a\in A}(x-a)=x^p-x \pmod{p}$$
which means that we can write $\prod_{a\in A}(x-a)=x^p-x+pF(x)$ for some polynomial $F(x)\in \mathbb Z[x]$. Since we have $w^p-w=0$, we can conclude that $$p^{-1}\prod_{a\in A}(w-a)=F(w)$$
so it is equal to a polynomial in $w$ with integer coefficients.
Next, let's look at what happens when we have $A=\{a_1,a_2,\dots,a_{p^k}\}$ which forms a complete set of residues modulo $p^k$. The polynomial $x^p-x$ has $p$ distinct roots modulo $p^k$, let's call them $B_0=\{b_1,b_2,\dots,b_p\}\subset A$. Let's define $B_r$ as the subset of $A$ which coincides with $\{b_1+r, b_2+r, \dots, b_p+r\}\pmod{p^k}$. Then we can write $A$ as a disjoint union:
$$A=B_0\cup B_p \cup B_{2p}\cup\cdots \cup B_{p^k-p}.$$
Suppose that $\ell_r$ is defined as the exponent of $p$ in the prime factorization of $r$, i.e. $p^{\ell_r}|| r$. Then we have
$$\prod_{a\in B_r} (x-a)=(x-r)^p-(x-r)+p^k G_r(x)=x^p-x+p^{\ell_r} H_r(x)$$
for some polynomials $G_r,H_r$ with integer coefficients. Therefore we can calculate by grouping
$$(p^k!)^{-1}\prod_{a\in A}(w-a)=\frac{1}{p^k!}\prod_{i=0}^{p^{k-1}-1}\prod_{a\in B_{pi}}(w-a)=\frac{1}{p^k!}(p^kG_0(w))\prod_{i=1}^{p^{k-1}-1}p^{\ell_{pi}}H_{pi}(w)$$
and this is $p$-integral because the exponent of $p$ in $p^k\cdot p^{\ell_p+\ell_{2p}+\cdots+\ell_{p^k-p}}$ matches that of $p$ in $p^k!$.
In order to finish the general case in the problem, let $n$ be written in base $p$ as $n_kp^k+n_{k-1}p^{k-1}+\cdots+n_0$ and then split the set $\{0,1,\dots,n-1\}$ into $n_k$ complete sets of residues modulo $p^k$, $n_{k-1}$ complete sets of residues modulo $p^{k-1}$ and so on, and then apply the argument from the previous paragraph to each of them.
