Here is an answer $\mathbf{if}$ $1\in X$:
Denote by $H$ the subgroup of $G$ being generated by $X$.
What you want is to show that $H=X^n$, which amounts to see that any element of $H$ can be written as a product of at most $n$ elements of $X$.
This follows from the fact that the size of $H$ is at most $n$:
We write an element $h$ of $H$ as a product of $m$ elements of $X$: $h=x_mx_{m-1}\dots x_1$ then we look at the elements $h_i=x_{i}\dots x_{1}$ obtained by the products of $i$ elements only. If $m>n$, there are two $h_i$'s which are equal, say $h_a=h_b$ with $a>b$. We replace $h_a$ with $h_b$ and can write $h$ with less elements.
$\mathbf{Edit}:$ If $1\notin X$, then $X^n$ is maybe not the group generated by $X$ (take for example the case where $X$ is a single element), but is in fact a subgroup, as Ilya showed.
