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 $(n + 1)! - 1$ .