This is not a complete solution but it seems to me that it can be a good starting point. I will prove that the system has $n!$ distinct solutions for a *general* choice of the coefficients $m_i$. 

All your equations are homogeneous in the $x_i$, except the last one. Homogenize adding a variable $x_0$; then all equations remain unchanged except for the last one, which becomes $$x_1x_2 \cdots x_n-x_0^n=0.$$
Now you can see the solutions of your system as the intersections of $n$ hypersurfaces in $\mathbb{P}^n(\mathbb{C})$, with homogeneous coordinates $[x_0: \ldots :x_n]$. If these hypersurfaces intersect transversally, then Bézout Theorem implies that there are only a finite number of solutions in the projective space, and this number is the product of the degrees of the equations, namely $1\cdot 2\cdot 3 \cdots \ n= n!$.

On the other hand, for $m_1=m_2 = \ldots =m_n=1$ you find exactly $n!$ solutions. So you can conclude that for a *general* choice of the coefficients $(m_1, \ldots, m_n) \in \mathbb{C}^n$ your system has exactly $n!$ solutions in $\mathbb{P}^n(\mathbb{C})$, all distinct.

Finally, these solution will be solutions of the original system if and only if all of them are outside the hyperplane at infinity $x_0=0$. This means that for any solution all the coordinates $x_1,\ldots, x_n$ must be nonzero. Again, you know that this happens in the case $m_1=m_2 = \ldots =m_n=1.$

So we can conclude that your system has exactly $n!$ solutions for a general choice of the coefficients $(m_1, \ldots, m_n) \in \mathbb{C}^n$, where *general* means that the coefficients can be chosen outside a Zariski closed subset $Z$ of $\mathbb{C}^n$. 

In particular the general $(m_1, \ldots, m_n) \in \mathbb{N}^n$ does not belong to $Z$, hence for a general choice of the positive integers $(m_1, \ldots, m_n)$ your system has precisely $n!$ distinct solutions.