# $\sum_{i=1}^n i\cdot i! = (n+1)!-1$ By Induction

I am trying to prove the following by Mathematical Induction:

$$\sum_{i=1}^n i\cdot i! = (n+1)!-1\quad\text{for all integers n\ge 1}$$

My proof by Induction follows:
First prove $P(1)$ is true,

$$\sum_{i=1}^1 i\cdot i! = (1+1)! – 1$$

Then, for all integers $k >= 1$, if $P(k)$ is true then $P(k+1)$ is also true,

$$\sum_{i=1}^k i\cdot i! = (k+1)! – 1$$
Then, we must show that $P(k+1)$ is true

$$\sum_{i=1}^{k+1} i\cdot i! = ((k+1)+1)! – 1\\ \sum_{i=1}^{k+1} i\cdot i! = (k+2)! – 1$$

I am currently struggling with the next step to show that $P(k)$ is equal to $P(k+1)$. How would I finish this prove by induction?

#### Solutions Collecting From Web of "$\sum_{i=1}^n i\cdot i! = (n+1)!-1$ By Induction"

Hint: $$\sum_{i=1}^{k+1} i\cdot i! = (k+1)\cdot(k+1)! + \sum_{i=1}^{k} i\cdot i!$$

by the definition of the $\sum$ operator.

Also, by the induction hypothesis, you know that $\sum_{i=1}^{k} i\cdot i!$, which appears on the right-hand side, is equal to $(k+1)!-1$.

If you want an alternative to induction, use telescoping sums:
$$\sum_{i=1}^ni\cdot i!=\sum_{i=1}^n(i+1-1)\cdot i!=\sum_{i=1}^n(i+1)\cdot i!-1\cdot i!$$
$$=\sum_{i=1}^n (i+1)!-i!=(n+1)!-n!+n!-\ldots-2!+2!-1!=(n+1)!-1.$$