Proving $\sum_{k=1}^n kk!=(n+1)!−1$

Question

Ok, so I'm trying to prove statement in the header. I have read the following discussion on it, but I can't seem to follow it all the way through:

Proving $\sum_{k=1}^n k k!=(n+1)!-1$

I like mfl's answer, but I get hung up on the last step. They say:

and we need to show

$$\sum_{k=1}^{n+1} kk!=(n+2)!−1.$$

Just write

$$\sum_{k=1}^{n+1} kk!=\sum_{k=1}^n kk! + (n+1)(n+1)!$$

How do they get from the first step stated above, to the following step? I'm stuck.

Answer 1

$$\sum_{k=1}^n kk!= \sum_{k=1}^{n-1} kk!+nn!$$

And by induction we know $\sum_{k=1}^{n-1} kk!= n!-1$

Combining we get: $$\sum_{k=1}^{n-1} kk!+nn! =n!-1+nn! = n!(1+n)-1=(n+1)!-1$$

Answer 2

Another way is to write : $$\sum_{k=1}^n k k!=\sum_{k=1}^n (k+1 -1) k!=\sum_{k=1}^n (k+1)!-k!=(n+1)!-1$$