Understanding a variant on the multinomial theorem in a commutative ring with unity

This post concerns Chapter 1 section "The Multinomial Theorem" on pages 65-67 of Analysis I by Amann and Escher.

Excerpts from text:

The part that I can't understand is the equation with the summation in this excerpt. The multinomial theorem (stated below) was proven immediately prior. Notation:

In case the notation isn't clear, we have the multi-index $$\alpha = (\alpha_1, \dots, \alpha_m) \in \mathbb N^m$$, and its length is $$\lvert \alpha \rvert := \sum_{j = 1}^m \alpha_j$$. We have $$\alpha ! := \prod_{j = 1}^m (\alpha_j)!$$. We also have $$a^{\alpha} := \prod_{j = 1}^m (a_j)^{\alpha_j}$$.

I am assuming that $$1 = 1_R$$ in the equation I don't understand. I have trouble explaining to myself why the form of the sum in the equation I don't understand is different from the form of the sum (on the right-hand side) in the multinomial theorem (8.4).

The first sentence of the proof is not hard to understand. However, the second sentence doesn't make sense to me. I'm sorry I can't be more specific. I guess I would ask why we need this special form (the equation I don't understand) when in the multinomial theorem, any of the $$a_j$$ could be equal to $$1$$ anyway? I can't reconcile the two.

I appreciate any help.

• The second sentence after the proof of Remark 8.6(a) doesn't belong to the proof, it's a separate remark about noncommutative rings (assuming that the given quantities do commute anyway). Aug 23 '20 at 17:35
• @Berci When I wrote "the second sentence" I was referring to the second sentence of the proof. The one that goes "The claim now follows from Theorem 8.5." Aug 23 '20 at 18:17

Exactly as you say, theorem 8.5 can be applied to elements $$(a_1,\dots,a_m,a_{m+1})$$ where $$a_{m+1}=1=1_R$$.
And this is what happens in the proof of remark 8.6(a). Let $$b$$ denote the sequence $$(a_1,\dots,a_m,1)$$ and note that $$b^\beta=a^\alpha$$ for the initial segment $$\alpha:=(\beta_1,\dots,\beta_m)$$ of an exponent sequence $$\beta=(\beta_1,\dots,\beta_m,\beta_{m+1})$$ because $$1^{\beta_{m+1}}=1$$.
Note also that the initial segment $$\alpha$$ uniquely determines the last term, hence the whole sequence $$\beta$$ if we assume $$|\beta|=k$$: specifically $$\beta_{m+1}$$ must be then $$k-|\alpha|$$, and $$\alpha$$ can be any sequence of exponents with $$\ |\alpha|\,\le \,k$$.
Using this correspondence we get $$(1+a_1+\dots+a_m)^k \ =\ \sum_{|\beta|=k} \frac{k!}{\beta!}b^\beta\ =\\ =\ \sum_{\matrix{|\alpha|\,\le\,k \\ \beta:=(\alpha_1,\dots,\alpha_m,k-|\alpha|)}} \frac{k!}{\beta!}b^\beta\ =\ \sum_{|\alpha|\le k} \binom k\alpha a^\alpha\ \ \ .$$