Proof By Induction - Factorials
$\left(\forall n \in \mathbb{N}\right)\left((n + 1)! = (n + 1) \cdot n!\right)$
Prove the following statement by induction: for all $n \in \mathbb{N}$
$\sum_{k=0}^{n}(k \cdot k!) = (n + 1)! − 1$
Base case: $n =0$
$(0\cdot 0!) = 0 ~\wedge~ (0+1)! - 1 = 0$, true.
Assume $n$ is true so $k = n+1$ $(n+1)\cdot(n+1)!= (n+1)\cdot(n+1)n!$
I'm not sure where to carry on from here? Can anyone shed some light?
Solution 1:
$$\sum_{k=0}^{n+1} k{\cdot}k!=\sum_{k=0}^nk{\cdot}k!+(n{+}1){\cdot}(n{+}1)!=(n{+}1)!{-}1+(n{+}1){\cdot}(n{+}1)!=(n{+}2){\cdot}(n{+}1)!{-}1=(n{+}2)!{-}1$$