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: {{begin-eqn}} {{eqn | l = \dbinom {k_1 + k_2} {k_1, k_2}       | r = \frac {\left({k_1 + k_2}\right)!} {k_1! \, k_2!}       | c = Definition of Multinomial Coefficient }} {{eqn | r = \frac {\left({k_1 + k_2}\right)!} {k_1! \, \left({\left({k_1 + k_2}\right) - k_1}\right)!}       | c =  }} {{eqn | r = \dbinom {k_1 + k_2} {k_1}       | c = Definition of Binomial Coefficient }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | r = \dbinom {k_1 + k_2 + \cdots + k_r + k_{r + 1} } {k_1, k_2, \ldots, k_r, k_{r + 1} }       | o =  }} {{eqn | r = \frac {\left({k_1 + k_2 + \cdots + k_r + k_{r + 1} }\right)!} {k_1! \, k_2! \, \cdots k_r! \, k_{r + 1}!}       | c =Definition of Multinomial Coefficient }} {{eqn | r = \frac {\left({k_1 + k_2 + \cdots + k_r}\right)!} {k_1! \, k_2! \, \cdots k_r!} \frac {\left({k_1 + k_2 + \cdots + k_r + k_{r + 1} }\right)!} {\left({k_1 + k_2 + \cdots + k_r}\right)! k_{r + 1} }       | c =  multiplying top and bottom by $\left({k_1 + k_2 + \cdots + k_r + k_{r + 1} }\right)!$ }} {{eqn | r = \dbinom {k_1 + k_2 + \cdots + k_r} {k_1, k_2, \ldots, k_r} \frac {\left({k_1 + k_2 + \cdots + k_r + k_{r + 1} }\right)!} {\left({k_1 + k_2 + \cdots + k_r}\right)! k_{r + 1} }       | c = Definition of Multinomial Coefficient }} {{eqn | r = \dbinom {k_1 + k_2 + \cdots + k_r} {k_1, k_2, \ldots, k_r} \dbinom {k_1 + k_2 + \cdots + k_r + k_{r + 1} } {k_1 + k_2 + \cdots + k_r}       | c = Definition of Binomial Coefficient }} {{eqn | 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} } \dbinom {k_1 + k_2 + \cdots + k_r + k_{r + 1} } {k_1 + k_2 + \cdots + k_r}       | c = Induction Hypothesis }} {{eqn | 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_{r + 1} } {k_1 + k_2 + \cdots + k_r} }} {{end-eqn}} So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall m \in \Z_{>1}: \dbinom {k_1 + k_2 + \cdots + k_m} {k_1, k_2, \ldots, k_m} = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2} \cdots \dbinom {k_1 + k_2 + \cdots + k_m} {k_1 + k_2 + \cdots + k_{m - 1} }$ {{qed}}