How to prove $\sum_{\alpha_1=0}^k \sum_{|\alpha|=k} \dfrac{k!}{\alpha!} a^\alpha = \sum_{|\alpha|=k} \dfrac{k!}{\alpha!} a^\alpha$

Question

In order to prove the multinomial theorem, I have to prove an intermediate result and have been stuck at the last step.

For $k,m \in \Bbb N$ with $m \ge 2$, $\alpha = (\alpha_1,\cdots,\alpha_{m+1}) \in \Bbb N^{m+1}$, $\alpha' = (\alpha_2,\cdots,\alpha_{m+1}) \in \Bbb N^{m}$, and $a=(a_1,\cdots,a_{m+1}) \in \Bbb R^{m+1}$, we define $$\begin{aligned}|\alpha| &:= \sum_{j=1}^{m+1}a_j\\ \alpha! &:= \prod_{j=1}^{m+1} (a_j)! \\ a^\alpha &:= \prod_{j=1}^{m+1}(a_j)^{\alpha_j}\\ {k \choose \alpha} &:= \frac{k!}{\alpha!(k-|\alpha|)!}\end{aligned}$$

Prove that $$\sum_{\alpha_1=0}^k {k \choose \alpha_1} a_1^{\alpha_1} \sum_{|\alpha'|=k-\alpha_1} \frac{(k-\alpha_1)!}{\alpha'!} a_2^{\alpha_2} \cdot \cdots \cdot a_{m+1}^{\alpha_{m+1}} = \sum_{|\alpha|=k} \frac{k!}{\alpha!} a^\alpha$$

My attempt:

We first have some observations.

1. $a_1^{\alpha_1} \cdot \cdots \cdot a_{m+1}^{\alpha_{m+1}} = a^\alpha$

2. $(\alpha_1)!\alpha'!=\alpha!$

3. $|\alpha'|=k-\alpha_1 \iff |\alpha|=k$

$$\begin{aligned}LHS &= \sum_{\alpha_1=0}^k \sum_{|\alpha'|=k-\alpha_1} {k \choose \alpha_1} a_1^{\alpha_1} \frac{(k-\alpha_1)!}{\alpha'!} a_2^{\alpha_2} \cdot \cdots \cdot a_{m+1}^{\alpha_{m+1}}\\ &= \sum_{\alpha_1=0}^k \sum_{|\alpha'|=k-\alpha_1} \frac{k!}{(\alpha_1)!(k-\alpha_1)!} \frac{(k-\alpha_1)!}{\alpha'!} a_1^{\alpha_1} \cdot \cdots \cdot a_{m+1}^{\alpha_{m+1}}\\ &= \sum_{\alpha_1=0}^k \sum_{|\alpha'|=k-\alpha_1} \dfrac{k!}{\alpha!} a^\alpha \text{ (from 1. and 2.)}\\ &= \sum_{\alpha_1=0}^k \sum_{|\alpha|=k} \dfrac{k!}{\alpha!} a^\alpha \text{ (from 3.)} \end{aligned}$$

From here, I am stuck at removing $\sum_{\alpha_1=0}^k$ to get the $RHS$. Intuitively, $\sum_{\alpha_1=0}^k \sum_{|\alpha|=k} \dfrac{k!}{\alpha!} a^\alpha = \sum_{|\alpha|=k} \dfrac{k!}{\alpha!} a^\alpha$, but I am unable to prove it rigorously.

Please help me prove this equality. Thank you for your help.

Answer 1

You can use this set equality \begin{align*} \big\{\alpha~:~|\alpha|=k\big\} = \bigcup_{i=0}^k \big\{\alpha~:~|\alpha|=k, \alpha_1=i\big\} \end{align*} and the fact that the sets on the right hand side are disjoints.