Prove by induction that $\sum_{k=0}^n k\cdot k! = (n+1)!-1$

Question

Prove by induction that $$\sum_{k=0}^n k\cdot k! = (n+1)!-1$$

I cannot manage to proceed after assuming that the equality works for $p$.

Answer 1

Suppose the desired equality holds for some fixed $n$. We'll show it holds for $n+1$.

$$ \begin{aligned} \sum_{k=0}^{n+1}k\cdot k! &=\sum_{k=0}^{n}k\cdot k! + (n+1)\cdot(n+1)!\\ &=(n+1)!-1+(n+1)\cdot(n+1)!\\ &=(1+n+1)(n+1)!-1\\ &=(n+2)\cdot(n+1)!-1\\ &=(n+2)!-1. \end{aligned} $$

Answer 2

First, show that this is true for $n=1$:

$\sum\limits_{k=0}^{1}k\cdot k!=(1+1)!-1$

Second, assume that this is true for $n$:

$\sum\limits_{k=0}^{n}k\cdot k!=(n+1)!-1$

Third, prove that this is true for $n+1$:

$\sum\limits_{k=0}^{n+1}k\cdot k!=$

$\color\red{\sum\limits_{k=0}^{n}k\cdot k!}+(n+1)\cdot(n+1)!=$

$\color\red{(n+1)!-1}+(n+1)\cdot(n+1)!=$

$(n+1)!\cdot(n+2)-1=$

$(n+2)!-1=$

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Please note that the assumption is used only in the part marked red.