For every finite subset I of N, let φ(I)=supJ⊂I∥pJ−e∥. For every subset J of I, one has
∥pJ−e−sJ∥≤φ(I)σJ.
The lemma is evident if J is empty; let us prove it by induction on the number of elements of J. Let J=K∪{j}, where j is strictly upper to all the elements of K; then pJ=pK(e+aj) and sJ=sK+aj, whence
pJ−e−sJ=(pK−e−sK)+(pK−e)aj
and, by virtue of the induction hypothesis and the definition of φ(I)
∥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.