Find a closed form for

Find a closed form for

S_n=1*1!+2*2!+...+n*n!.

for integers n>=1. Your response should have a factorial.

$\begin{array}{|rcll|} \hline n &=& (n+1) - 1 \\ \hline n\cdot n! &=& [~(n+1) - 1~]\cdot n! \\ n\cdot n! &=& (n+1)\cdot n! - n! \\ \mathbf{n\cdot n!} &\mathbf{=}& \mathbf{(n+1)! - n!} \\ \hline \end{array}$

$\begin{array}{|r|r|c|r|} \hline n & n\cdot n! &=& (n+1)! - n! \\ \hline 1 & 1* 1! &=& \not \! 2! - 1! \\ 2 & 2* 2! &=& \not \! 3! - \not \! 2! \\ 3 & 3* 3! &=& \not \! 4! - \not \! 3! \\ 4 & 4* 4! &=& \not \! 5! - \not \! 4! \\ 5 & 5* 5! &=& \not \! 6! - \not \! 5! \\ \dots & \dots & & \dots \\ n & n\cdot n! &=& (n+1)! - \not \! n! \\ \hline \Sigma & 1*1!+2*2!+...+n*n! &=& (n+1)! - 1! \\ \hline \end{array}$

Find a closed form for

S_n=1*1!+2*2!+...+n*n!.

for integers n>=1. Your response should have a factorial.

(1!+2 2!+...+n n!) = sum_(k=1)^n k k! = (n+1)!-1

Find a closed form for

S_n=1*1!+2*2!+...+n*n!.

for integers n>=1.

Your response should have a factorial.

Example for $n=4$:

$\begin{array}{|rcll|} \hline S_4 &=& 1*1!+2*2!+3*3!+4*4! \quad & | \quad 4! = 3!*4 \\ S_4 &=& 1*1!+2*2!+3*3!+4*3!*4 \\ S_4 &=& 1*1!+2*2!+3!*(3+4*4) \\ S_4 &=& 1*1!+2*2!+3!*(3+4^2) \quad & | \quad 3! = 2!*3 \\ S_4 &=& 1*1!+2*2!+ 2!*3*(3+4^2) \\ S_4 &=& 1*1!+2!*[~2+ 3*(3+4^2)~] \quad & | \quad 2! = 1!*2 \\ S_4 &=& 1*1!+1!*2*[~2+ 3*(3+4^2)~] \\ S_4 &=& 1!*\{~1+2*[~2+ 3*(3+4^2)~]~\}\quad & | \quad 1! = 1 \\ S_4 &=& 1*\{~1+2*[~2+ 3*( \underbrace{3+4^2}_{ \underbrace{=4-1+4^2}_{ \underbrace{=4*(4+1)-1}_{=4*5-1} } })~]~\} \\ S_4 &=& 1*\{~1+2*[~2+ 3*( 4*5-1 )~]~\} \\ S_4 &=& 1*[~1+2*(~2+ 3*4*5-3 ~)~] \\ S_4 &=& 1*[~1+2*(~3*4*5-1 ~)~] \\ S_4 &=& 1*(~1+2*3*4*5-2 ~) \\ S_4 &=& 1*(~ 2*3*4*5-1 ~) \\ S_4 &=& 1*2*3*4*5-1\\ S_4 &=& 5!-1\\ 1*1!+2*2!+3*3!+4*4! &=& 5!-1\\ \hline \end{array}$

$\begin{array}{|rcll|} \hline S_n &=& 1*1!+2*2!+...+n*n! \\ S_n &=& (n+1)!-1 \\ \hline \end{array}$

Or as follows:

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