Let $n \geq 2$, Let $(Q,A)$ be an n-group, $e$ its {1,n}-neutral operation [:[6], 1.3], and $^{-1}$ its inversing operation [:[7], 1.3]. Let also $Z$ be an set of all integers. Then, in this paper, we say that $a^m (m \in Z)$ is an m-th power of the element $a$ in $(Q,A)$ iff: (1) $a^1=a$ (def); (2) $a^{k+1}=A(a^k, \overset{n-2,a},a),k \geq 1$ (def); (3) $a^0=e(\overset{n-2,a})$ (def) and (4) $a^{-k}=(\overset{n-2,a},a^k)^{-1},k \geq 1$ (def) [:2.3]. Furthermore, for all $a \in Q$ and for all $s \in Z$ the following equality holds: $a^{\langle s \rangle}=a^{s+1}$, where $$a^{angle s \rangle}$ well-known is the s-th n-adic power of the element $a$ in $(Q,A)$ [:2.5]; ${angle s \rangle}=s(n-1)+1$. Among others, in the paper is proved the following proposition. For every $lpha, lpha_1, \dots, lpha_n ı Z$ $(n \geq 3)$ the following equalities hold: $e(a^{lpha_1}, \dots, a^{lpha_{n-2}})=a^{-um_{i=1}^{n-2}lpha_i+n-2}$ , $(a^{lpha_1}, \dots, a^{lpha_{n-2}}, a^{lpha})^{-1}=a^{-lpha-2(um_{i=1}^{n-2}lpha_i-n+2)}$ and $A(a^{lpha_1}, \dots, a^{lpha_n})=a^{um_{i=1}^{n}lpha_i-n+2}$ [:2.7,2.8,footnote 4)].