Editing Multinomial theorem (section)

===Proof===
This proof of the multinomial theorem uses the [[binomial theorem]] and [[Mathematical induction|induction]] on {{mvar|m}}.

First, for {{math|1=''m'' = 1}}, both sides equal {{math|''x''{{sub|1}}{{sup|''n''}}}} since there is only one term {{math|1=''k''{{sub|1}} = ''n''}} in the sum. For the induction step, suppose the multinomial theorem holds for {{mvar|m}}.  Then

: <math>
\begin{align}
& (x_1+x_2+\cdots+x_m+x_{m+1})^n = (x_1+x_2+\cdots+(x_m+x_{m+1}))^n \\[6pt]
= {} & \sum_{k_1+k_2+\cdots+k_{m-1}+K=n}{n\choose k_1,k_2,\ldots,k_{m-1},K} x_1^{k_1} x_2^{k_2}\cdots x_{m-1}^{k_{m-1}}(x_m+x_{m+1})^K
\end{align}
</math>

by the induction hypothesis.  Applying the binomial theorem to the last factor,

:<math> = \sum_{k_1+k_2+\cdots+k_{m-1}+K=n}{n\choose k_1,k_2,\ldots,k_{m-1},K} x_1^{k_1}x_2^{k_2}\cdots x_{m-1}^{k_{m-1}}\sum_{k_m+k_{m+1}=K}{K\choose k_m,k_{m+1}}x_m^{k_m}x_{m+1}^{k_{m+1}}</math>
:<math> = \sum_{k_1+k_2+\cdots+k_{m-1}+k_m+k_{m+1}=n}{n\choose k_1,k_2,\ldots,k_{m-1},k_m,k_{m+1}} x_1^{k_1}x_2^{k_2}\cdots x_{m-1}^{k_{m-1}}x_m^{k_m}x_{m+1}^{k_{m+1}}
</math>

which completes the induction.  The last step follows because

:<math>{n\choose k_1,k_2,\ldots,k_{m-1},K}{K\choose k_m,k_{m+1}} = {n\choose k_1,k_2,\ldots,k_{m-1},k_m,k_{m+1}},</math>
as can easily be seen by writing the three coefficients using factorials as follows:

:<math> \frac{n!}{k_1! k_2! \cdots k_{m-1}!K!} \frac{K!}{k_m! k_{m+1}!}=\frac{n!}{k_1! k_2! \cdots k_{m+1}!}.</math>