Multinomial Coefficient expressed as Product of Binomial Coefficients

Theorem

 * $\dbinom {k_1 + k_2 + \cdots + k_m} {k_1, k_2, \ldots, k_m} = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2} \cdots \dbinom {k_1 + k_2 + \cdots + k_m} {k_1 + k_2 + \cdots + k_{m - 1} }$

where:
 * $\dbinom {k_1 + k_2 + \cdots + k_m} {k_1, k_2, \ldots, k_m}$ denotes a multinomial coefficient
 * $\dbinom {k_1 + k_2} {k_1}$ etc. denotes binomial coefficients.

Proof
The proof proceeds by induction.

For all $m \in \Z_{> 1}$, let $P \left({m}\right)$ be the proposition:
 * $\dbinom {k_1 + k_2 + \cdots + k_m} {k_1, k_2, \ldots, k_m} = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2} \cdots \dbinom {k_1 + k_2 + \cdots + k_m} {k_1 + k_2 + \cdots + k_{m - 1} }$

Basis for the Induction
$P \left({2}\right)$ is the case:

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:
 * $\dbinom {k_1 + k_2 + \cdots + k_r} {k_1, k_2, \ldots, k_r} = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2} \cdots \dbinom {k_1 + k_2 + \cdots + k_r} {k_1 + k_2 + \cdots + k_{r - 1} }$

from which it is to be shown that:
 * $\dbinom {k_1 + k_2 + \cdots + k_r + k_{r + 1} } {k_1, k_2, \ldots, k_r, k_{r + 1} } = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2} \cdots \dbinom {k_1 + k_2 + \cdots + k_r + k_{r + 1} } {k_1 + k_2 + \cdots + k_r}$

Induction Step
This is the induction step:

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

Therefore:
 * $\forall m \in \Z_{>1}: \dbinom {k_1 + k_2 + \cdots + k_m} {k_1, k_2, \ldots, k_m} = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2} \cdots \dbinom {k_1 + k_2 + \cdots + k_m} {k_1 + k_2 + \cdots + k_{m - 1} }$