Choose finite symmetric generating sets

S_{i}

for all

G_{i}

, and let

L \geq 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}

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}

G_{i}

, each of

S_{i - 1}

-length at most

r

, and hence

S

-length at most

L r

The

(r + 1)^{n}

products

t_{1, j_{1}} t_{2, j_{2}} \dots t_{n, j_{n}} (0 \leq j_{i} \leq 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

n L r

. Therefore

(r + 1)^{n} \leq ∣ B_{S} (n L r) ∣ \leq C (1 + n L r)^{d} .

Letting

r \to \infty

yields

n \leq d .

For

n = 0

the assertion is automatic.

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