I was trying to prove the following.
$$(x_1+x_2+\cdots+x_m)^n =\sum_{a_1+\cdots+a_m=n}\binom{n}{a_1,...,a_m} x_1^{a_1}\cdots x_m^{a_m}$$
I'll appreciate it if someone reviews my work.
Suppose we have at most m categories so that we can assign the n elements of a set (namely $B$), to those categories. Lets say I have the set $C=\{x_1,x_2,...x_m\}$ of my m categories, and $a_i$ is the size of my category. So $a_i \geq 0$ for all i. I'll associate to C the generating function $x_1+x_2+\cdots + x_m$. If i take $H= C \times C \times \cdots \times C$ (n times), then an element of H is of the form $(x_1,x_1,x_3,...,x_m)$, for example. The number of entries that are equal to $x_i$ will tell me the number of elements of $B$ that are assigned into that category. The generating function of H would then be $(x_1+x_2+\cdots + x_m)^n$. If we expand that product, then we would get:
$$\sum_{a_1+\cdots+a_m=n} \prod_{x_i \in C} x_i^{a_i}$$
It then can be seen that:
$$\sum_{a_1+\cdots+a_m=n} \prod_{x_i \in C} x_i^{a_i} = \sum_{a_1+\cdots+a_m=n}\binom{n}{a_1,...,a_m} x_1^{a_1}\cdots x_m^{a_m} $$
I am not quite satisfied with the last inference. I would like it if someone gives a more rigorous answer. Also, let me know any mistakes. I am just starting with generating functions.