by using the definition of permutation we can get an equivalent form

$1+1\times 1!+2\times 2!+3\times 3!+\cdots+n\times n! $

now , $r\times r! =r\times r! +r!-r!= (r+1)r!-r!$

now, $1\times 1!=2!-1!,$

and $2\times 2!=3!-2!$

and so on

$n\times n!=(n+1)!-n! $

the first equation becomes,

$1+2!-1!+3!-2!+4!-3!+\cdots+(n+1)!-n! = (n+1)!$

by converting the terms into permutation definition we get above result