Let $m_1,m_2,\dots,m_n\in \mathbb{Z}_0^{+}$ such that $\sum_{i=1}^nm_i\geq {n \choose 2}$. Let us write $m=\sum_{i=1}^nm_i-{n \choose 2}$. Prove that the coefficient of $(x_1+x_2+\dots+x_n)^m \prod_{1\leq j<i<n}(x_i-x_j)$ in the monomial $x_1^{m_1}x_2^{m_2}\cdots x_n^{m_n}$ is $$\frac{m!}{m_1! m_2!\cdots m_n!}\prod_{1\leq j<i\leq n}(m_i-m_j).$$
I noticed that the $$f(x_1,x_2,\dots ,x_n)=\prod_{1\leq j<i<n}(x_i-x_j)$$ is the vandermonde polynomial, and I guess that $\deg(f)={n \choose 2}$ since $f$ is non zero if all the $x_i\neq x_j$.
Now for the other side I have $$g(x_1,x_2,\dots, x_n)=(x_1+x_2+\cdots+x_n)^m$$ which is pretty simetric and also I guess that a coefficient of $g$ in any monomial is $${m \choose k_1,k_2,\dots, k_n}x_1^{k_1}x_1^{k_2}\cdots x_n^{k_n}.$$
Now I try paste both parts but I can't figure out how I should do it.
I think that I should think how should be the $k_i$ since $\deg (g)=m$ and for get the monomial $x_1^{m_1}x_2^{m_2}\cdots x_n^{m_n}$ I need a monomial of $f$ which degree is ${n \choose 2}$ and complete the sum of $m_i$.
Edit:
I try do a particular case when $n=3$ and $m_1=3,m_2=1,m_3=1$ and search the monomial $x_1^3x_2x_3$ and then I expand the polynomial $$(x_1-x_2)(x_2-x_3)(x_1-x_3)(x_1+x_2+x_3)^2=$$ $$x_2 x_1^4 - x_3 x_1^4 + x_2^2 x_1^3 - x_3^2 x_1^3 - x_2^3 x_1^2 + x_3^3 x_1^2 - x_2^4 x_1 + x_3^4 x_1 - x_2 x_3^4 - x_2^2 x_3^3 + x_2^3 x_3^2 + x_2^4 x_3$$ and first I try watch a regular expression or a pattern, but it seems like there aren't pattern and second I don't get the term and $x_1^3x_2x_3$ in the expression above, now check my calculations and are right, but in these case I don't know where I made the mistake?.
Also I try only expand the vandermonde polynomial for $3$ variables.
$$(x_3-x_1)(x_2-x_1)(x_3-x_2)=x_3^2(x_2-x_1)+x_2^2(x_1-x_3)+x_1^2(x_3-x_2)$$
And I think that for $4$ variables I should obtain $$x_3^3(x_2-x_1)+x_2^3(x_1-x_3)+x_1^3(x_3-x_2)+x^4(x_4-x_1)$$ but it is wrong, and I don't identify a pattern.
Any comment, hint or suggestion is useful. Thanks in advance.