Fractional Binomial Coefficients I recently examined the binomial coefficient $\binom{\frac{1}{2}}{k}$ and found that the denominator was always a power of two. The same is true of $\binom{\frac{1}{3}}{k}$, where the denominator is always a power of three. While the first proof was simple, the case of $\frac{1}{3}$ was messy and involved counting the powers of 3 in the numerator and denominator. The case of $\frac{1}{p}$ for prime $p$ can be done in the same bashy way. Is it possible there is a more elegant proof?
 A: Here’s a purely conceptual proof, more advanced, that has the advantage of involving no computation and no induction:


*

*Let $N$ be a positive number, and let $P_N(t)$ be the polynomial function that tells you the coefficient of $x^N$ in $(1+x)^t$. You know that $P_N$ has rational coefficients, and takes integer values when $t$ is evaluated to a positive integer.

*Now consider any prime number $q$. The polynomial $P_N$ has its coefficients in the field ${\mathbb{Q}}_q$ of $q$-adic numbers, is continuous, and takes values in the $q$-adic integers ${\mathbb{Z}}_q$ whenever $t$ is evaluated to a positive integer.

*But any rational number without $q$ in its denominator is a $q$-adic integer, so $q$-adically approximable by ordinary integers, and in fact by positive integers. Thus if $\alpha$ is a rational with no $q$ in its denomiator, we get $P_N(\alpha)\in{\mathbb{Z}}_q$, by continuity of $P_N$.

*In particular, the coefficients of $(1+x)^{1/p}$ are rational numbers that are in ${\mathbb{Z}}_q$ for each $q\ne p$: the only denominators are powers of $p$.


A more advanced argument allows you to say just how divisible by $p$ the denominators of the coefficients of $(1+x)^{1/p}$ will be, but that’s a story for another day.
A: Using the identity $$\binom{\frac{1}{p}}{k}=\frac{(-1)^k}{k!}\prod_{j=0}^{k-1} \left({j-\frac{1}{p}}\right)$$ I found a fairly simple proof for why the denominator of this binomial coefficient is always a power of $p$.
Let $p^a$ be the highest power of $p$ that divides $k!$. Then, $$\prod_{j=0}^{k-1}\left({j-\frac{1}{p}}\right) \equiv \prod_{j=0}^{k-1}({j+n}) \equiv 0 \pmod{ \frac{k!}{p^a}}$$ for some $n \equiv \frac{1}{p} \pmod {\frac{k!}{p^a}}$, with congruence to 0 resulting from the fact that $k!$ always divides a product of $k$ consecutive integers. Thus we have $$\prod_{j=0}^{k-1}\left({j-\frac{1}{p}}\right)=\left(\frac{k!}{p^a}\right)z$$ for $z \in \mathbb{Z}$. Substituting then provides,
$$\binom{\frac{1}{p}}{k}=\frac{(-1)^k}{k!}\left(\frac{k!}{p^a}\right)z=\frac{(-1)^kz}{p^a}$$ completing the proof.
A: Observe
\begin{align}
(1 + x)^{1/p} = 1 + \frac{x}{p} - \frac{1-p}{2p^{2}} x^{2} + \frac{1 - 3p + 2p^{2}}{6 p^{3}} x^{3} - \cdots
\end{align}
A: Firstly, note the definition of binomial coefficients with arbitrary real upper co-efficient:
$${r\choose k}=\begin{cases}\frac{r^{\underline{k}}}{k!} & k\in\mathbb{Z}^{*} \\ 0 & k\in\mathbb{Z}^{-}\end{cases}$$
Where $r^{\underline{k}}=r(r-1)(r-2)\cdots(r-k+1)$ is the falling factorial.
Now for your problem, let $r=\frac{1}{x},\forall x\in\mathbb{N}$. It is fairly evident that the bottom will contain at least one power of $x$, whilst the top will contain no powers of $x$, therefore $x$ will divide the denominator of ${r \choose k}$. Which is what you are trying to prove.
