Since we have Φn(F0,...,Fn)≡Φn(X0p,...,Xnp)mod.pn+1. A for every n∈N (formulas (2) and (15)), it follows from prop. 1, b) that we have Fn≡Xnpmod.p. A for every n∈N.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.