= Solution
Choose finite symmetric generating sets $S_i$ for all $G_i$, and let $L\ge1$ 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 $t_{i,0},\ldots,t_{i,r}$ of $G_i$, each of $S_{i-1}$-length at most $r$, and hence $S$-length at most $Lr$.
The $(r+1)^n$ products
$$
t_{1,j_1}t_{2,j_2}\cdots t_{n,j_n}\qquad(0\le j_i\le r)
$$
are 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 $nLr$. Therefore
$$
(r+1)^n\le|B_S(nLr)|\le C(1+nLr)^d.
$$
Letting $r\to\infty$ yields
$$
\boxed{n\le d.}
$$
For $n=0$ the assertion is automatic.
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