Letu=∑aνXν∈A[[I]]andman integer>0. For everyn∈Nlet(αν,n)ν∈N(I)be the family of coefficients ofun. Ifα0m=0, thenαν,n=0forn≥∣ν∣+m.
Let ν∈N(I) and n∈N. We have
αν,n=ν(1)+⋯+ν(n)=ν∑αν(1)⋯αν(n).
If n≥∣ν∣+m and ν(1)+⋯+ν(n)=v, we have ∣ν(1)∣+⋯+∣ν(n)∣≤n−m. We thus have v(r)=0 and so αv(r)=ar, for at least m distinct values of r; it follows that αν(1)⋯αν(n)=0, whence the result.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.