To show that $n!f(x) \in \mathbb{Z}[x]$ [closed]

If $f(x)$ is a polynomial such that if $y \in\mathbb{Z}$, then $f(y)\in\mathbb{Z}$, show that there exist $n$ such that $n!f(x)\in\mathbb{Z}[x]$

closed as off-topic by Watson, Joey Zou, Claude Leibovici, JonMark Perry, ervxAug 6 '16 at 13:22

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• What's $n$? Obviously if you choose $n$ large enough you can clear all coefficients' denominators. Is perhaps $n$ the degree of your polynomial? – Gregory Grant Jul 22 '16 at 13:36
• And where if $f$ taken from? $\Bbb Q [x]$, $\Bbb R[x]$, $\Bbb C[x]$? – Alex M. Jul 22 '16 at 13:37

A well known result is that any integer-valued polynomial is the sum of integer multiples of $\binom{x}{k} =\dfrac{x(x-1)...(x-k+1)}{k!}$.
By multiplying by the largest $k!$, we get your statement.
Newton's interpolation formula is $$f(n) = d_0 \binom{n}{0} + d_1 \binom{n}{1} + d_2 \binom{n}{2} + d_3 \binom{n}{3} +\cdots$$ where $d_i$ are the numbers in the first column of the repeated differences array.