Factorial Summation Problem

Question

$$\sum_{j=0}^n j\cdot j!$$

I got $(n+1)!-1$ as the answer but I'm not sure if that's right or how I even got to that answer exactly. (my paper is a mess of random work and I can't make it out). Can somebody tell me if that's correct and explain the best and condensed way to get to the answer?

Answer 1

Hint: Write $j \cdot j! = (j+1)j! - j! = (j+1)!-j!$. Now you have a telescoping sum.

Alternatively, you can prove that statement using mathematical induction.

Answer 2

Try a telescoping sum, like so $$(n+1)! - 0! = \sum_{k-1}^n \left( (k+1)! - k!\right).$$ Can you bring it home?