TITLE: Polynomial in $\mathbb{Q}[x]$ sending integers to integers?
QUESTION [14 upvotes]: We can view the binomial coefficient $\binom{x}{k}$ has a polynomial in $x$ with degree $k$. So taking some $f\in\mathbb{Q}[x]$, why is $f(n)\in\mathbb{Z}$ for all $n\in\mathbb{Z}$, precisely when the coefficients of $f$ are integers with respect to the basis $\{\binom{x}{k}\mid k\in\mathbb{N}\}$?
The reverse direction seems clear, so I wonder why the opposite implication is also true. I did notice that all the roots of $\binom{x}{k}$ are just $0,\dots,k-1$. Can we factor the polynomial is some nice way to reveal the opposite conclusion? Thanks.

REPLY [3 votes]: Since you already know that the $\binom Xn$ for $n\in\Bbb N$ form a $\Bbb Q$-basis of $\Bbb Q[X]$ (and indeed this is easy to see), the rest of the argument is trivial. Let $P=\sum_nc_n\binom Xn\in\Bbb Q[X]$ be arbitrary, then the value of $P$ at $m\in\Bbb N$ is $c_m+\sum_{i<n}c_i\binom mi$, which must be an integer. Now by immediate induction on $m$, all the coefficients $c_m$ are integer.