TITLE: Combinatorial Proof of Multinomial Theorem - Without Induction or Binomial Theorem
QUESTION [8 upvotes]: I've been trying to rout out an exclusively combinatorial proof of the Multinomial Theorem with bounteous details but only lighted upon this one - see P2. Any other helpful ones?

$(x_1+\cdots+x_k)^n =$ Top of Page 39 from UNC:
  $\sum\limits_{\large{(a_1, ..., a_{k - 1}) \; \ni \; 0 \le a_1+...+ a_{k - 1} \le n}}\dbinom{n}{a_1,...,a_{k-1}}\cdot x_1^{a_1} \cdots x_{k-1}^{a_{k-1}}x_k^{\large{n - a_1 - \cdots - a_{k-1}}}$
  $= \sum\limits_{\large{a_1+...+a_k = n \; \& \; a_i \ge 0 }}\dbinom{n}{a_1,...,a_k} x_1^{a_1} \cdots x_k^{a_k}$ Bottom of P1 from MSU.   

I see that both are the Multinomial Theorem but which one is better? 
I thought to try this myself, following the combinatorial proof of the Binomial Theorem. So  esteem each term (in green) in $(x_1+\cdots+x_k)^n = \underbrace{\color{green}{[x_1+\cdots+x_k]...[x_1+\cdots+x_k]}}_{\text{n terms}}$ as one box from which to choose $x_1, ..., x_k$.
Since each term/box (in green) contains $k$ terms, the total number of terms $ = k^n$.
First, consider $x_1$.
● For $x_1^n$, must have $a_ 1 = n\qquad \& \qquad a_2 = ... = a_k = 0$.
● For $x_1^{n - 1}$, must have $a_1 = n - 1 \qquad \& \qquad a_2 \text{ OR } ... \text{ OR } a_k = 1$
(the latter due to $a_1+...+a_k = n$ in definition from P1 of MSU)
...
● For $x_1^{1},$ must have $a_1 = 1 \qquad \& \qquad a_2 \text{ OR } ... \text{ OR } a_k = n - 1$
● For $x_1^{0},$ must have $a_1 = 0 \qquad \& \qquad a_2 \text{ OR } ... \text{ OR } a_k = n$
The above extends to and must hold for all $x_1, ..., x_k$.
How to translate all this into combinatorial notation? How to forge ahead and complete please?

REPLY [2 votes]: The multinomial theorem says that for commuting quantities $X_1,\ldots,X_n$ in a ring one has
$$
  (X_1+\cdots+X_k)^n=\sum_{\substack{a_1,\ldots,a_k\in\Bbb N\\a_1+\cdots+a_k=n}}
  \binom n{a_1,\ldots,a_k}X_1^{a_1}\ldots X_k^{a_k}
$$
where the multinomial coefficient $\binom n{a_1,\ldots,a_k}$ can be defined (under the hypothesis $a_1+\cdots+a_k=n$, without which it is either left undefined, or maybe set to$~0$), either combinatorially as the number of words (of length$~n$) over the alphabet $\{A_1,\ldots,A_k\}$ containing $a_i$ copies of the letter$~A_i$ for $i=1,\ldots,k$, or algebraically as $\frac{n!}{a_1!\ldots a_k!}$.
A combinatorial proof of the multinomial theorem would naturally use the combinatorial description of multinomial coefficients. It is almost a triviality: without assuming that the $X_i$ commute, the $n$-th power can be expanded by the distributive laws to a sum of $k^n$ distinct monomials in the $X_i$, which are just all distinct words of length $n$ that can be formed from them; now assuming commutation, one groups together all words that have the same number of copies of each$~X_i$, and if those multiplicities are $a_1,\ldots,a_k$ respectively, then one gets $\binom n{a_1,\ldots,a_k}$ times the monomial $X_1^{a_1}\ldots X_k^{a_k}$.
The equivalence of the combinatorial and algebraic description of multinomial coefficients is another matter, but also quite simple. For a word of length$~n$ with given multiplicities $a_1,\ldots,a_k$ of individual letters, the $n!$ permutations of the letters of the word certainly generate all words that can be made from that multiset of letters, but each word is obtained multiple times. To obtain from one way to generate a word all other ways to generate the same word, one can independently permute all copies of a same letter among each other, which can be done in $a_1!\ldots a_k!$ ways. This means there are$\frac{n!}{a_1!\ldots a_k!}$ distinct words.