I know that a factorized polynomial can be written in the following way:
$$(x_{11}+...+x_{1n})^{a_1}...(x_{n1}+...+x_{nn})^{a_n}=\sum_{k_{11}+...+k_{1n}=a_1\\\quad\quad\vdots\\k_{n1}+...+k_{nn}=a_n\\}\binom{a_1}{k_{11},...,k_{1n}}...\binom{a_n}{k_{n1},...,k_{nn}}x_{11}^{k_{11}}x_{12}^{k_{12}}...x_{nn}^{k_{nn}}$$
While solving another problem I stubled upon a slightly different sum:
\begin{align*} &\sum_{k_{11}+...+k_{1n}=a_1\\\quad\quad\vdots\\k_{n1}+...+k_{nn}=a_n\\}\frac{(k_{11}+...+k_{n1})!}{k_{11}!...k_{1n}!}...\frac{(k_{1n}+...+k_{nn})!}{k_{n1}!...k_{nn}!}x_{11}^{k_{11}}x_{12}^{k_{12}}...x_{nn}^{k_{nn}} \\ =&\sum_{k_{11}+...+k_{1n}=a_1\\\quad\quad\vdots\\k_{n1}+...+k_{nn}=a_n\\}\binom{k_{11}+...+k_{n1}}{k_{11},...,k_{n1}}...\binom{k_{1n}+...+k_{nn}}{k_{1n},...,k_{nn}}x_{11}^{k_{11}}x_{12}^{k_{12}}...x_{nn}^{k_{nn}} \end{align*}
The difference consists in the "vertical" summation instead of the "horizontal" of the $k_{ij}$ in the multinomial coefficient. Are there still any possible ways to simplify this sum?