Definition
A topological group is a group $(X,\,*,\,\cal T)$ equipped with a topology $\cal T$ with respect to which the functions $$ p:X\times X\ni (x_1,x_2)\longrightarrow x_1*x_2\in X\quad\text{and}\quad s:X\ni x\longrightarrow x^{-1}\in X $$ are continuous.
So we say that a neighborhood $V_0$ of $x_0\in X$ is symmetric if the identity holds $$ V_0=V^{-1}_0 $$ where we put $$ V^{-1}_0:=\{x\in X:x=y^{-1}\,\text{with }y\in V_0\} $$
So we observe that if $x\in V_0^n$ for any $n\in\Bbb N^+$ then there exists $\xi_1(x),\dots,\xi_n(x)\in V_0$ such that $$ x=\xi_1(x)*\dots*\xi_n(x) $$ so that if $x\in(V_0^n)^{-1}$ and so $x^{-1}\in V_0^n$ then there exists $\xi_1(x^{-1}),\dots,\xi_n(x^{-1})\in V_0$ such that $$ x=\big(\xi_n(x^{-1})\big)^{-1}*\dots*\big(\xi_1(x^{-1})\big)^{-1} $$ but $V_0$ is symmetric and thus $\xi\big(\xi_n(x^{-1})\big)^{-1},\dots,\big(\xi_1(x^{-1})\big)^{-1}\in V_0$ which proves that $x\in V_0^n$: so we conclude that $$ (V_0^n)^{-1}\subseteq V_0^n $$ Finally, we observe that $$ x^{-1}=\big(\xi_1(x)*\dots*\xi_n(x)\big)^{-1}=\big(\xi_n(x))^{-1}*\dots*\big(\xi_1(x))^{-1} $$ for any $x\in V_0^n$ but again $V_0$ is symmetric and so $\big(\xi_1(x)\big)^{-1},\dots,\big(\xi_n(x)\big)^{-1}\in V_0$ so that $x^{-1}\in V_0^n$ and thus $$ x=(x^{-1})^{-1}\in(V_0^n)^{-1} $$ which proves that $$ V_0^n\subseteq(V_0^n)^{-1} $$
So I ask if the result effectively holds and thus I ask if I well proved it or if it is possibile to prove it with more simple argumentations. So coulds someone help me, please?
This has nothing to do with topology, let alone neighborhoods (I corrected the title of this post.) You simply proved that if a subset $V$ of a group satisfies $V=V^{-1}$, then $V^n=(V^n)^{-1}$.
(Notice in passing that $V\subset V^{-1}$ and $V^{-1}\subset V$ are equivalent, hence both equivalent to $V=V^{-1}$.)
Your proof can be written more succintly: since (for any subset $V$) $(V^n)^{-1}=(V^{-1})^n$, $$V=V^{-1}\Rightarrow V^n=(V^{-1})^n=(V^n)^{-1}.$$