Let $r \geq 1$; the Følner sequences $\{F^{n,1}\}_{n=1}^{\infty}, \{F^{n,2}\}_{n=1}^{\infty}, \dotsc, \{F^{n,r}\}_{n=1}^{\infty}$ satisfy the bounded intersection property if there is a constant p such that, for any $n\in\mathbb{N}$ and $1\leq i\leq r$, each $F^{n,i}$ can intersect no more than p disjoint translates of $F^{n,1}, F^{n,2},\dotsc, F^{n,r}$. They have comparable magnitudes if $0<\varliminf_{n}|F^{n,i}|/|F^{n,j}|<\infty$ for $1\leq i,j\leq r$. Suppose that G is a countable Abelian group with an element of infinite order, and let $\mathcal{X}$ be a mixing finite (or $\beta$-local, with $\beta>1/2$) rank action of G on a probability space. Suppose that the Følner sequences of $\mathcal{X}$ satisfy the bounded intersection property, and have comparable magnitudes. Then $\mathcal{X}$ is mixing of all orders. We follow Ryzhikov's joining technique in our proof: the main theorem follows from showing that any pairwise independent joining of k copies of $\mathcal{X}$ is necessarily product measure.