We use mathematical induction. In fhe basis step, for n = 1, the equation states that 1 · 1! = (1 + 1)! − 1, and this is true because both sides of the equation evaluate to 1. For the inductive step, we assume that 1 · 1! + 2 · 2! + · · · + k · k! = (k + 1)! − 1 for some positive integer k. We add (k + 1)(k + 1)! to the left hand side to find that 1 · 1! + 2 · 2! + · · · + (k + 1) · (k + 1)! = (k + 1)! − 1 + (k + 1)(k + 1)! The right hand side equals (k + 1)!(k + 2) − 1 = (k + 2)! − 1. This establishes the desired equation also for k + 1, and we are done by the principle of mathematical induction.We use mathematical induction. In fhe basis step, for n = 1, the equation states that 1 · 1! = (1 + 1)! − 1, and this is true because both sides of the equation evaluate to 1. For the inductive step, we assume that 1 · 1! + 2 · 2! + · · · + k · k! = (k + 1)! − 1 for some positive integer k. We add (k + 1)(k + 1)! to the left hand side to find that 1 · 1! + 2 · 2! + · · · + (k + 1) · (k + 1)! = (k + 1)! − 1 + (k + 1)(k + 1)! The right hand side equals (k + 1)!(k + 2) − 1 = (k + 2)! − 1. This establishes the desired equation also for k + 1, and we are done by the principle of mathematical induction.