How can we prove this = 1 for all n

Question

$\displaystyle n!-\sum_{k=1}^{n-1}k\cdot k!$

By computing this by hand for several small values of $n$ I can see that it is always equal to 1. But I can't see how to prove that.

Answer 1

Use induction. For $n=1$ this is trivial. If the result holds for $n$, then $$(n+1)!-\sum_{k=1}^n k\cdot k! = (n+1)n!-n\cdot n!-\sum_{k=1}^{n-1} k\cdot k!=n!-\sum_{k=1}^{n-1} k\cdot k!=1$$ thus the result holds for all positive integers.

Answer 2

Hint: $k\cdot k! = (k+1)! - k!$