How is $1+d+d(d-1)+\ldots+d(d-1)^{r-1}<d(d-1)^r$ in the following proof? Here $d$ is the maximum degree of a graph, $n$ is the number of vertices of a graph and $\delta$ is the diameter of a graph.
Theorem $\boldsymbol{2.6}$. Assume $d \geq 3$. Then we have $$\delta > \frac{\log n}{\log(d-1)} -2.$$ For a regular graph, we also have $$\gamma < 2 \frac{\log n}{\log(d-1)} +2.$$ Proof. Let $u$ be a fixed vertex. Then balls around $u$ grow at most exponentially with respect to the radius: $$|B_r(u)| \leq 1+d+d(d-1)+\ldots+d(d-1)^{r-1}<d(d-1)^r.$$
Hint: It is $$1+d+d(d-1)+\cdots +d(d-1)^{r-1}=1+{\frac {d \left( d-1 \right) ^{r}}{d-2}}-{\frac {d}{d-2}}$$