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

Question

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

I tried to use the method that is generally applied to geometric series, i.e. writing $S_n - \phi(\cdot)S_n = \dots$ in such a way that almost all terms cancel out. The problem is that in this case it does not work because there is nothing constant in the series' terms.

Answer 1

Hint
See that $$(n+2)! - (n+1)! = (n+1)(n+1)!$$ And use induction.

Answer 2

Let's prove by induction that $\sum\limits_{k=1}^{n}{k}\cdot{k!}=(n+1)!-1$.

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

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

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

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

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

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

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

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

• $(n+1)!\cdot(n+2)-1=$

• $(n+2)!-1$

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

Answer 3

Note that $(k+1)!=k!(k+1) \Rightarrow k!k=(k+1)!-k!$. Use this in your sum to find: $$ \sum_{k=1}^n k!k= (2!-1!)+(3!-2!)+(4!-3!)+\cdots + (n+1)!-n! $$ and eliminating opposite terms you have the result.