Let $\operatorname{Int}(\mathbb{Z}):= \{f(x) \in \mathbb{Q}[x] \mid f(\mathbb{Z}) \subseteq \mathbb{Z}\}$. Any element of the integer valued polynomials can be written as a $\mathbb{Z}$ linear combination of the polynomials of the form $f_n(x) = x(x-1)...(x-n+1)/n!$. As a consequence, we can show that $\operatorname{Int}(\mathbb{Z})$ is a free $\mathbb{Z}$- module.
Now I am stuck at a self-check that every finitely generated ideal of $\operatorname{Int}(\mathbb{Z})$ is invertible. I have tried to compute the explicit inverse ideal of $I = \langle f_1,...,f_n\rangle$ but have not succeeded. My idea in a nutshell was to first prove it for an ideal of the form of $I$, then do it for any integral linear combinations, but have failed. If you could provide me with some hints and suggestions I would be highly obliged.