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

Solution 1:

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 $(k+1)!-1$.

Solution 2:

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