Mathematical Induction Factorials, sum r(r!) =(n+1)! -1 How do I prove that $$\sum\limits_{r=1}^{n} r(r!) = (n+1)!-1$$
I was able to get to factor: $LHS = k(k!) + (k+1)(k+1)!$
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\, RHS = (k+2)! -1$
 A: $\newcommand{\+}{^{\dagger}}%
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$$
\color{#0000ff}{\large\sum_{r = 1}^{n}r\pars{r!}}
=
\sum_{r = 1}^{n}\bracks{\pars{r + 1}\pars{r!} - r!}
=
\sum_{r = 1}^{n}\bracks{\pars{r + 1}! - r!}
$$
$$
\color{#0000ff}{\large\sum_{r = 1}^{n}r\pars{r!}}
=
\pars{2! - 1!} + \pars{3! - 2!} + \cdots + \bracks{\pars{n + 1}! - n!}
=
\pars{n + 1}! - 1! = \color{#0000ff}{\large\pars{n + 1}! - 1}
$$
