p=0 and σp=(−1)ν(σ) for σ∈Sn.
p(1,2,…,n)=∏i<j(j−i)=0 and hence p=0. On the other hand, if σ∈Sn, then
σp(z1,…,zn)=p(zσ(1),…,zσ(n))=i<j∏(zσ(j)−zσ(i)).
Let C be the set of ordered pairs (i,j) such that 1≤i≤n, 1≤j≤n, i<j. A permutation θ is defined on C by setting θ(i,j)=(σ(i),σ(j)) if (i,j) is not an inversion, θ(i,j)=(σ(j),σ(i)) if (i,j) is an inversion. This implies σp=(−1)ν(σ)p.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.