1 + 1.1! + 2.2! + 3.3! +......+ n.n! is equal to

$\sum _{r=1}^n r\cdot r!=\sum _{r=1}^n \left(\right(r+1)!-r!$

$=\left(2!-1!\right)+\left(3!-2!\right)+......+\left(n!-\left(n-1\right)!\right)+\left(\left(n+1\right)!-n!\right)$

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

1 + 1.1! + 2.2! + 3.3! +......+ n.n! =(n+1)!