Find sum of series $n\ n!$

Question

I don't know where to start, all I am able to achieve is:

$\sum\limits_{n=1}^k n\ n! = \sum\limits_{n=1}^k n(n-1)! $

Thanks to Wolfram I know the solution and it's $(1 + k)! -1$, but I don't know how to get there.

Answer 1

To the OP. One not so difficult way to show this equality is to use induction. You can see it is true for $k=1$. Assume the equality is true for arbitrary $k$, all you need to show now is $(k+1)!-1+(k+1)(k+1)!=(k+2)!-1$, which is little algebra on the left side. From the terms $(k+1)!$ and $(k+1)(k+1)!$ you can factor $((k+1)!$ leaving you with $(k+1)!(1+k+1)$ with that $-1$ tagging along. Can you finish it?

Answer 2

$$n\,n!=(n+1)!-n!$$ so that you have a telescopic sum that equals the last term minus the first, $$(k+1)!-1!$$