algebra precalculus - Proof by induction: $sumlimits_{i=0}^n i cdot i! = (n+1)!-1$

Let n be a positive natural number , $n\ge 0$, then. $\displaystyle\sum_{i=0}^n i \cdot i!= (n+1)!-1$

Here is my attempt.I'm not going to write the base case because I understand that part.

Assuming $\displaystyle\sum_{i=0}^n i \cdot i!= (n+1)!-1$ is true. We wish to show.

$\displaystyle \sum_{i=0}^{n+1} i \cdot i!= (n+2)!-1$. Thus.

$\displaystyle\sum_{i=0}^{n+1} i \cdot i!= (\sum_{i=0}^n i \cdot i!) $ This is where I get stuck.

Answer

Your last step should read $$\sum_{i=0}^{k+1} \left( i \cdot i! \right)= \sum_{i=0}^k \left(i \cdot i!\right) + (k+1)(k+1)! $$ Now use your induction assumption to get $$\sum_{i=0}^{k+1} \left(i \cdot i! \right)= (k+1)! - 1 + (k+1)(k+1)! = (k+1)! (k+2) -1 = (k+2)! - 1$$