*Let x1,…,xs be elements of X such that s≥1 and xi=xi+1 for 1≤i≤s−1; let n1,…,ns be non-zero rational integers. Then the element ∏i=1s(1+xi)ni of A^(X) is =1.
Let m be a maximal ideal of K and k the field K/m; let p:A^K(X)→A^k(X) be the unique continuous homomorphism of unital K-algebras such that p(x)=x for x∈X (no. 1, Proposition 1). It suffices to prove that p(∏i=1s(1+xi)ni)=1 and the problem is reduced to the case where K is a field.
In the notation of Lemma 2:
i=1∏s(1+xi)ni=b1≥0∑⋯bs≥0∑cb1,n1⋯cbs,nsx1b1⋯xsbs.
By Lemma 2, there exist integers ai>0 such that cai,ni=0 (1≤i≤s). By Algebra, Chapter I, § 7, no. 4, Proposition 6, no monomial x1b1⋯xsbs such that bi≥0 and (b1,…,bs)=(a1,…,as) can be equal to x1a1…xsas. It follows that the coefficient of x1a1…xsas in ∏i=1s(1+xi)ni is ca1,n1…cas,ns=0, which implies the result.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.