Given a vector $\vec{v}\in\mathbb{Z}^n$ whose entries are relatively prime, is it possible to build an invertible $n\times n$ integer matrix which has $\vec{v}$ as a column?
For $n=2$ this is true, and it seems like it should be true for $n>2$, but the proof becomes much more complicated as the expression for the determinant becomes more complicated. I am more or less trying to set the determinant equal to $\pm 1$ and then determine if the corresponding equations have integer solutions. This is messy. I am wondering if there is a nicer way to prove this. The proof doesn't need to be elementary.
Yes; the Smith Normal Form algorithm is a sledgehammer that will do this.
More generally, one can transform a column $\pmatrix{a_1\\\vdots\\a_n}$ with integer entries and $\gcd(a_1,\ldots,a_n)=1$ to the unit vector $\pmatrix{1\\0\\\vdots\\0}$ via row operations of the form $R_i\mapsto R_i+mR_j$ ($i\ne j$, $n\in\Bbb Z$), $R_i\leftrightarrow R_j$ and $R_i\mapsto R_j$, each of which corresponds to premultiplying by an elementary matrix with integer entries and determinant $\pm1$. Thus there is a matrix $M$ integer entries and determinant $\pm1$ with $M\pmatrix{a_1\\\vdots\\a_n}=\pmatrix{1\\0\\\vdots\\0}$. Then $M^{-1}$ also has integer entries, and determinant $\pm1$ and its first column is $\pmatrix{a_1\\\vdots\\a_n}$.