Yeah it is obvious, since a determinant is just a polynomial of its terms so if all terms are integers, then obviously the determinant will be an integer. I wasn't thinking straight last night

It becomes clear when seeing the Leibniz formula for the determinant. We have that for an $\displaystyle n \times n$-matrix $\displaystyle A=(a_{ij})$, then

$\displaystyle \det(A) = \sum_{\sigma \in S^n} \mathrm{sgn}(\sigma) \prod_{i=1}^n a_{i,\sigma(i)},$

where $\displaystyle S^n$ is the symmetric group of $\displaystyle n$ elements.