(related to Proposition: Sum of Factorials (I))
We provide a proof by induction for all natural numbers $n.$ * Base case: $n=0$ $$\sum_{k=0}^0 k\cdot k !=0\cdot 0 !=0\cdot 1=1-1=(0+1)!-1.$$ * Induction step $n\to n+1$ * Assume, $$\sum_{k=0}^n k\cdot k !=(n+1) !-1$$ is correct for an $n\ge 0.$ * Then $$\begin{align}\sum_{k=0}^{n+1} k\cdot k !&=\sum_{k=0}^{n} k\cdot k ! + (n+1)\cdot (n+1) !\nonumber\\ &=(n+1)!-1 +(n+1)\cdot (n+1) !\nonumber\\ &=(n+2)!-1.\nonumber\end{align}$$ * Thus, the above sum of factorials holds for all natural numbers $n\in\mathbb N.$
