For each finite subset I of N, let φ(I)=supJ⊂I∣∣pJ−e∣∣. Then for each subset J of I we have
∣∣pJ−e−sJ∣∣≤φ(I)σJ.
The lemma is obvious if J is empty; we shall prove it by induction on the number of elements in J. Let J=K∪{j}, where {j} is strictly larger than every i∈K. Then pJ=pK(e+aj) and
sJ=sK+aj,
so that
pJ−e−sJ=pK−e−sK+(pK−e)aj,
and by the inductive hypothesis and the definition of φ(I), we have
∣∣pJ−e−sJ∣∣≤φ(I)σK+φ(I)∣∣aj∣∣=φ(I)σJ,
which proves the lemma.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.