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

Question

(whenever $n$ is a non-negative integer)

I did the basic step $P(1)$ and found the statment $P(n+1)$

I now have $(n+1)! - 1 + (n+1)\cdot(n+1)!$

This should equal $(n+2)! - 1$, but how do I show that?

Answer 1

You have two terms containing the expression $(n+1)!$. Put these together and you are done.

Answer 2

For $n=k$ , let $$\sum_{j=1}^{k}j\times j\,!=(k+1)!-1$$ If $n=k+1$ we have $$\sum_{j=1}^{k+1}j\times j\,!=(k+1)\times(k+1)!+\sum_{j=1}^{k}j\times j\,!=(k+1)\times(k+1)!+(k+1)!-1\\ \qquad=(k+1)!(k+1+1)-1=(k+1)!(k+2)-1=(k+2)!-1$$

Answer 3

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