Now

*I* am not sure what you mean by n-case and (n+1)-case. A proof by induction starts with identifying a property of natural numbers, often denoted by P(n). For each n, P(n) is either true or false (in particular, P(n) is not a number). This property can be an equality, as in the current problem, but it can have other forms: e.g., e

_{1}(n) divides e

_{2}(n) for some expressions e

_{1} and e

_{2}, e(n) is a

triangular number for some e, etc. The problem is to prove that P(n) holds for all natural n ≥ n

_{0}. The base case consists of proving P(n

_{0}). The induction step consists of proving that P(n) implies P(n+1) for all n ≥ n

_{0}.

In this problem, P(n) is 1(1!) + 2(2!) + ... + n(n!) = (n+1)! - 1; therefore, P(n+1) is 1(1!) + 2(2!) + ... + (n+1)(n+1)! = (n+2)! - 1. I assumed the line

1(1!) + 2(2!)+...+n(n!)(n+1) = ((n+1)+1))! - 1

in post #1 supposed to be P(n+1). That's why I said n(n!)(n+1) should be replaced by (n+1)(n!)(n+1) = (n+1)(n+1)!.