The proof proceeds by induction on $m$. For each $m \in \N_{\ge 1}$, let $\map P m$ be the proposition: :$\displaystyle \forall n \in \N: \paren {x_1 + x_2 + \cdots + x_m}^n = \sum_{k_1 \mathop + k_2 \mathop + \mathop \cdots \mathop + k_m \mathop = n} \binom n {k_1, k_2, \ldots, k_m} {x_1}^{k_1} {x_2}^{k_2} \cdots {x_m}^{k_m}$ === Basis for the Induction === Trivially, for all $n \in \N$: {{begin-eqn}} {{eqn | l = \paren {x_1}^n       | r = \sum_{k_1 \mathop = n} \frac {n!} {k_1!} {x_1}^{k_1}       | c =  }} {{eqn | r = \frac {n!} {n!} {x_1}^n       | c =  }} {{eqn | r = {x_1}^n       | c =  }} {{end-eqn}} and so it is seen that $\map P 1$ holds. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \forall n \in \N: \paren {x_1 + x_2 + \cdots + x_r}^n = \sum_{k_1 \mathop + k_2 \mathop + \mathop \cdots \mathop + k_r \mathop = n} \binom n {k_1, k_2, \ldots, k_r} {x_1}^{k_1} {x_2}^{k_2} \cdots {x_r}^{k_r}$ from which it is to be shown that: :$\displaystyle \forall n \in \N: \paren {x_1 + x_2 + \cdots + x_r + x_{r + 1} }^n = \sum_{k_1 \mathop + k_2 \mathop + \mathop \cdots \mathop + k_r \mathop + k_{r + 1} \mathop = n} \binom n {k_1, k_2, \ldots, k_r, k_{r + 1} } {x_1}^{k_1} {x_2}^{k_2} \cdots {x_r}^{k_r}  {x_{r + 1} }^{k_{r + 1} }$ === Induction Step === {{begin-eqn}} {{eqn | o =       | r = \paren {x_1 + x_2  + \cdots + x_r + x_{r + 1} }^n }} {{eqn | r = \paren {\paren {x_1 + x_2  + \cdots + x_r} + x_{r + 1} }^n }} {{eqn | r = \sum_{j \mathop = 0}^n \binom n j {x_{r + 1} }^j \paren {x_1 + x_2 + \cdots + x_r}^{n - j}       | c = Binomial Theorem }} {{eqn | r = \sum_{j \mathop = 0}^n \binom n j {x_{r + 1} }^j \sum_{k_1 + k_2 + \cdots + k_r \mathop = n - j} \binom {n - j} {k_1, k_2, \ldots, k_r} {x_1}^{k_1} {x_2}^{k_2} \cdots {x_r}^{k_r}        | c = Induction Hypothesis }} {{eqn | r = \sum_{j \mathop = 0}^n \paren {\sum_{k_1 + k_2 + \cdots + k_r \mathop = n - j} \binom n j \binom {n - j} {k_1, k_2, \ldots, k_r} {x_1}^{k_1} {x_2}^{k_2} \cdots {x_r}^{k_r} {x_{r + 1} }^j}       | c = Summation is Linear }} {{eqn | r = \sum_{k_{r + 1} \mathop = 0}^n \paren {\sum_{k_1 + k_2 + \cdots + k_r \mathop = n - k_{r + 1} } \binom n {k_{r + 1} } \binom {n - k_{r + 1} } {k_1, k_2, \ldots, k_r} {x_1}^{k_1} {x_2}^{k_2} \cdots {x_r}^{k_r} {x_{r + 1} }^k_{r + 1} }       | c = renaming index variable }} {{eqn | r = \sum_{k_1 + k_2 + \cdots + k_r + k_{r + 1} \mathop = n} \binom n {k_{r + 1} } \binom {n - k_{r + 1} } {k_1, k_2, \ldots, k_r} {x_1}^{k_1} {x_2}^{k_2} \cdots {x_r}^{k_r} {x_{r + 1} }^{k_{r + 1} }       | c = collapsing the double sum }} {{end-eqn}} Now: {{begin-eqn}} {{eqn | l = \binom n {k_{r + 1} } \binom {n - k_{r + 1} } {k_1, k_2, \ldots, k_r}       | r = \frac {n!} {k_{r + 1}! \paren {n - k_{r + 1} }!} \frac {\paren {n - k_{r + 1} }!} {k_1! \, k_2! \, \cdots k_r!}       | c = {{Defof|Binomial Coefficient}} and {{Defof|Multinomial Coefficient}} }} {{eqn | r = \dfrac {n!} {k_1! \, k_2! \, \cdots k_r! \, k_{r + 1}!} }} {{eqn | r = \binom n {k_1, k_2, \ldots, k_r, k_{r + 1} }       | c = {{Defof|Multinomial Coefficient}} }} {{end-eqn}} So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction. {{qed}}