1. **State the problem:** Show that $k \times k! = (k + 1)! - k!$ and then use this to sum the series $1 \times 1! + 2 \times 2! + 3 \times 3! + \cdots + n \times n!$. 2. **Show the formula and explain:** Recall the factorial definition: $k! = k \times (k-1) \times \cdots \times 1$. 3. **Prove the identity:** Start with the right side: $$ (k+1)! - k! = (k+1) \times k! - k! = k! \times ((k+1) - 1) = k! \times k = k \times k!. $$ This confirms the identity. 4. **Sum the series:** The series is: $$ S = \sum_{k=1}^n k \times k!. $$ Using the identity, rewrite each term: $$ k \times k! = (k+1)! - k!. $$ So, $$ S = \sum_{k=1}^n ((k+1)! - k!) = \sum_{k=1}^n (k+1)! - \sum_{k=1}^n k!. $$ 5. **Simplify the sums:** Write out terms: $$ \sum_{k=1}^n (k+1)! = 2! + 3! + 4! + \cdots + (n+1)!, $$ $$ \sum_{k=1}^n k! = 1! + 2! + 3! + \cdots + n!. $$ Subtracting, $$ S = (2! + 3! + \cdots + (n+1)!) - (1! + 2! + 3! + \cdots + n!) = (n+1)! - 1!. $$ 6. **Final answer:** $$ \boxed{\sum_{k=1}^n k \times k! = (n+1)! - 1}. $$ This shows the sum of the series in a simple closed form.