Thank you for these solutions to Shahnawaz Abdullah; Daniel, Liceo Scientifico Copernico, Torino, Italy; Anderthan, Saratoga High School; Andrei, School 205, Bucharest, Romania; David, Queen Mary's Grammar School, Walsall; Paddy, Peter, Greshams School, Holt, Norfolk; Ngoc Tran, Nguyen Truong To High School (Vietnam); Chris, St. Bees School; Dorothy, Madras College; A Ji and Hyeyoun, St. Paul's Girls' School; and Yatir from Israel.

To prove that k.k! = (k+1)! - k!.

If we take k! out as a factor from the right hand side of the equation, we are left with k!((k+1)-1) which simplifies to k.k!, as required.

Now we sum the series 1.1!+.....n.n!

As we have proved, n.n! is equal to (n+1)! - n! and therefore (n-1).(n-1)! is equal to (n-1+1)! - (n-1)! which simplifies to n! - (n-1)!. If we add the two results, we find that n! cancels.

If we sum the series from 1 to n, we find that all of the terms cancel except for (n+1)! and -(1!). Thus the sum of all numbers of the form r.r! from 1 to n is equal to (n+1)! - 1.