Having More Fun with Factorials

It can be shown by induction that $1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + 4 \cdot 4! +... + n \cdot n! = (n+1)! - 1$ Then our desired expression equals $11*11! - 1$ We can evaluate this by hand using the fact that $10! = 3628800$ or use a calculator and see the sum of the digits is 53.

Using calculator gives $\displaystyle \sum_{i=1}^{10} x\times x!=39916799$

So we got the answer $3+9+9+1+6+7+9+9=53$