Multinomial Theorem

Theorem
Let $x_1, x_2, \ldots, x_k \in F$, where $F$ is a field.

Then:


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

where:
 * $m \in \Z_{> 0}$ is a positive integer
 * $n \in \Z_{\ge 0}$ is a non-negative integer
 * $\dbinom n {k_1, k_2, \ldots, k_m} = \dfrac {n!} {k_1! \, k_2! \, \cdots k_m!}$ denotes a multinomial coefficient.

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 it shall be considered to have a value of $1$.

The multinomial theorem is a generalization of the Binomial Theorem.

Proof
The proof proceeds by induction on $m$.

For each $m \in \N_{\ge 1}$, let $P_m$ be the proposition:


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

Basis for the Induction
Trivially, for all $n \in \N$:

and so it is seen that $P_1$ holds.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 2$, then it logically follows that $P \left({r + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\displaystyle \forall n \in \N: \left({x_1 + x_2 + \cdots + x_r}\right)^n = \sum_{k_1 \mathop + k_2 \mathop + \mathop \cdots \mathop + k_r \mathop = n} \binom n {k_1, k_2, \ldots, k_r} {x_1}^{k_1} {x_2}^{k_2} \cdots {x_r}^{k_r}$

from which it is to be shown that:
 * $\displaystyle \forall n \in \N: \left({x_1 + x_2 + \cdots + x_r + x_{r + 1} }\right)^n = \sum_{k_1 \mathop + k_2 \mathop + \mathop \cdots \mathop + k_r \mathop + k_{r + 1} \mathop = n} \binom n {k_1, k_2, \ldots, k_r, k_{r + 1} } {x_1}^{k_1} {x_2}^{k_2} \cdots {x_r}^{k_r} {x_{r + 1} }^{k_{r + 1} }$

Induction Step
Now:

So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction.