= Solution
Choose a finite symmetric generating set $S$ for $\Gamma$, a base point $p$, and put $L=\max_{s\in S}d(p,sp)$. If the <word length> of $\gamma$ is at most $m$, the <triangle inequality> and invariance of <Riemannian distance> under <isometries> give $d(p,\gamma p)\leq Lm$.
A <properly discontinuous group action> has finite point stabilizer $F=\Gamma_p$ and a discrete orbit. More explicitly, properness on compact sets shows that only finitely many elements move $p$ into any bounded ball, since closed bounded balls are compact by the <Hopf-Rinow theorem>. Thus there is $\rho>0$ such that distinct orbit points have distance more than $2\rho$: choose it using the finitely many orbit points within a fixed small ball of $p$, and translate the resulting separation by <isometries>. If the whole orbit is one point, the group itself is finite and the conclusion is immediate.
Let $N_m$ be the number of distinct orbit points represented by the <word metric> ball of radius $m$. Their radius-$\rho$ balls are disjoint, have equal positive volume $v_0=\operatorname{Vol}B(p,\rho)$, and lie inside $B(p,Lm+\rho)$. The allowed <Bishop-Gromov inequality> for nonnegative <Ricci curvature> states that $\operatorname{Vol}B(p,R)/R^n$ is nonincreasing for $R>0$. Therefore
$$
N_mv_0\leq\operatorname{Vol}B(p,Lm+\rho)\leq v_0\left(\frac{Lm+\rho}{\rho}\right)^n.
$$
Each orbit point has at most $|F|$ preimages inside the word ball, since its full fiber is a coset of $F$. For the <growth function of a finitely generated group>, this gives
$$
\boxed{\beta_\Gamma(m)\leq |F|\left(1+\frac{Lm}{\rho}\right)^n=O(m^n).}
$$
Hence $\Gamma$ has <polynomial growth of a group> of degree at most $n$. The finite stabilizer factor makes the proof valid even when the properly discontinuous action is not free. This is <polynomial growth of properly discontinuous isometry groups>.
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