Let $G$ be a group of order $n> 1$ and let $\{x_1, \ldots, x_{d(G)}\}$ be a minimal set of generators for $G$. Taken, however, an automorphism $f$ of $G$, then it is in particular an application bijective fixing the unit of $G$, therefore, for $f(x_1)$ there are $n-1$ choices, as the order of $G \ {1}$. Choose $f(x_1)$, for $f(x_2)$ there are $| G \setminus \langle {f(x_1 )}\rangle |$ choices because, not being $x_2$ in $\langle{x_1}\rangle$, then $f(x_2) \not\in \langle{f(x_1 )}\rangle$: so the choices for $f(x_2)$ are $n-2$ because, for the exercise above, $|\langle {f(x_1 )}\rangle | \geq 2$. Assigned values to $f(x_1)$ and $f(x_2)$, from the exercise above follows that for $f(x_3)$ there are $|G \setminus \langle{f(x_1),f(x_2 )}\rangle | \leq (n-2^2)$ choices. So proceeding, for $f(x_{d(G)})$ remain at most $(n-2^{d(G) -1})$ choices. Then $|\operatorname{Aut}(G)| \leq \prod_{i = 0}^{d (G) -1} (n-2)^i$. In conclusion, observing that $d(G) \leq log_2 (n)$ then $d(G) -1 < log_2〖(n)〗$, or $2 ^{d(G) -1} < n$; therefore $2^{d(G) -1} \leq n-1$ and thus $d(G) -1 \leq log_2(n-1)$. Then $0 \leq i \leq d(G) -1 \leq \lfloor log_2(n-1) \rfloor$ and exercise is proven.