Consider the divided-power ring $A := \mathbb Z \langle x_1, \ldots, x_n \rangle$ consisting of $\mathbb Z$-linear combinations of divided-power monomials of the form $x_1^{(a_1)} \cdots x_n^{(a_n)}$; this can be defined as the subring of the polynomial ring $\mathbb Q[ x_1, \ldots, x_n ]$ which is generated as a $\mathbb Z$-algebra by the elements $ x_i^{(m)} := \displaystyle \frac 1 {m!} x_i^m $.

Next consider the lattice $V \subseteq A$ generated by $x_1, \ldots, x_n$; that is, $V$ is the subspace of $A$ consisting of degree-1 polynomials. Then divided powers of elements of $V$ are elements of $\mathbb Q[ x_1, \ldots, x_n ]$, and one can check that they are in fact in $A$. Now my question is the following: Do the divided powers of elements of $V$ generate $A$ as an abelian group? (This is motivated by the fact that for a field $k$, the divided-power ring $k\langle x_1, \ldots, x_n \rangle := A \otimes_{\mathbb Z} k$ is spanned as a $k$-vector space by the divided powers of degree-1 polynomials). EDIT: Following Scott Carnahan's answer below, I need to take $k$ algebraically closed here.