# Prove that $\sum_{i=1}^{n} i \times i! = (n+1)! - 1$ by induction

\begin{align*} \sum_{i = 1}^{k + 1} i(i!) & = \sum_{i = 1}^{k} i(i!) + (k + 1)(k + 1)!\\ & = (k + 1)! - 1 + (k + 1)(k + 1)! & \text{by the induction hypothesis}\\ & = (1 + k + 1)(k + 1)! - 1\\ & = (k + 2)(k + 1)! - 1\\ & = (k + 2)! - 1 \end{align*}

I have a question from this post solving the problem Prove by induction that $\sum_{i=1}^n i!\times i=(n+1)!-1$ for all $n\in \mathbb{N}$

How does the person go from $$= (k + 2)(k + 1)! - 1\\ = (k + 2)! - 1$$

at the very end? I don't understand how the permutation of $$(k+1)!$$ and (k+2) are able to combine into $$(k+2)!$$

• Consider an example, say with $k=3$: $$(k+2)(k+1)!=(3+2)(3+1)!=5\cdot4!=5(4\cdot3\cdot2\cdot1)=5\cdot4\cdot3\cdot2\cdot1=5!=(3+2)!=(k+2)!$$ Feb 6, 2019 at 15:16
• ah that makes sense, thank you Feb 6, 2019 at 22:51

It is because, by definition, $$n! = n(n-1)!$$ and $$0! = 1$$. Just take $$n=k+2$$.
$$\begin{eqnarray*} \sum_{i=1}^{n} i\cdot i! & = & \sum_{i=1}^{n} (i+1-1)\cdot i! \\ & = & \sum_{i=1}^{n} \left((i+1)!- i!\right) \\ & \stackrel{telescoping}{=} & (n+1)! - 1\\ \end{eqnarray*}$$
• The OP might wonder how you got from $(i+1)i!$ to $(i+1)!$.... Feb 6, 2019 at 15:13