 Solution

Choose finite symmetric generating sets $S_i$ for all $G_i$, and let $L\ge 1$ bound the $S$-length of their elements. If $[G_{i-1}:G_i]=\infty$, the connected locally finite Schreier graph of left cosets of $G_i$ in $G_{i-1}$ is infinite. Its radius-$r$ ball has at least $r+1$ vertices: otherwise a finite-diameter locally finite connected graph would be finite. Equivalently, we can choose $r+1$ distinct left-coset representatives $t_{i,0},\dots ,t_{i,r}$ of $G_i$, each of $S_{i-1}$-length at most $r$, and hence $S$-length at most $Lr$.

The $(r+1{)}^n$ productsare distinct. Indeed, equality first forces equality of the $G_1$-cosets of the first factors, and hence their indices; cancellation then forces equality of the $G_2$-cosets of the second factors, and so on. Each product has $S$-length at most $nLr$. ThereforeLetting $r\to \infty$ yieldsFor $n=0$ the assertion is automatic.