Multinomial Theorem

Theorem
Let
 * $\displaystyle \binom n {k_1, k_2, \ldots, k_m} = \frac{n!}{k_1!k_2!\cdots k_m!}$

be the multinomial coefficient. Let $x_1, x_2, \ldots, x_k\in F$, where $F$ is a field and $\exists i\ni x_i\ne 0$. Then


 * $\displaystyle \left({x_1 + x_2 + \cdots + x_m}\right)^n = \sum_{k_1 + k_2 + \cdots + k_m \mathop = n} \binom n {k_1, k_2, \ldots, k_m} x_1^{k_1} x_2^{k_2} \cdots x_m^{k_m},$

where $m$ is a positive integer and $n$ is a non-negative integer.

The sum is taken for all non-negative integers $k_1, k_2, \ldots, k_m$ such that $k_1 + k_2 + \cdots + k_m = n$, and with the understanding that wherever $0^0$ may appear on the right-hand side it shall be considered to have a value of $1$.

The multinomial theorem is the generalization of the binomial theorem.

Proof
The proof proceeds by induction on $m$:

For each $m\in \mathbb{N}_1$, let $P_m$ be the following proposition:


 * $\displaystyle \forall n \in \N: \left({x_1 + x_2 + \cdots + x_m}\right)^n = \sum_{k_1+k_2+\cdots+k_m \mathop = n} \binom n {k_1, k_2, \ldots, k_m} x_1^{k_1} x_2^{k_2} \cdots x_m^{k_m}$

Base Case ($m = 1$)
Trivially, for all $n \in \N$:


 * $\displaystyle \left({x_1}\right)^n = \sum_{k_1 \mathop = n} \frac{n!}{k_1!} x_1^{k_1} = \frac{n!}{n!} x_1^n = x_1^n,$

so $P_1$ holds.

Induction Step
Suppose that $P_m$ holds. Then

Now,

Therefore, we have demonstrated that for $m\in\mathbb{N}_1$, $P_m\implies P_{m+1},$

thus completing the induction step.

The result follows by mathematical induction.