Let (1+x)n=∑r=0nar.xr.  then   (1+a1a0)(1+a2a1)−−−−−(1+anan−1)  is equal to

(n+1)c1nc0.(n+1)c2nc1−−−−−(n+1)cnncr−1 =1(n+1)c1nc0.2.(n+1)c2nc1−−−−−(n+1).nc2ncr−1.1n! =(n+1)nc0nc0(n+1)nc1nc1−−−−−(n+1).ncr−1ncr−1.1n!=(n+1)nn!