Thank you for these solutions to Shahnawaz
Abdullah;  Daniel Disegni, Liceo Scientifico
Copernico, Torino, Italy; Anderthan Hsieh,
Saratoga High School; Andrei Lazanu, School 205,
Bucharest, Romania; David Moxey, Queen Mary's
Grammar School, Walsall; Paddy Snow; Peter
Barton, Greshams School, Holt, Norfolk;  Ngoc
Tran, Nguyen Truong To High School (Vietnam);
Chris Tynan, St. Bees School; Dorothy Winn,
Madras College; A Ji Yang; Hyeyoun Chung, St.
Paul's Girls' School; Yatir Haslevi.
<p/>
To prove that <i>k</i>.<i>k</i>! = (<i>k</i>+1)! &#8722; <i>k</i>!.
<p/>
If we take <i>k</i>! out as a factor from the right
hand side of the equation, we are left with
<i>k</i>!((<i>k</i>+1)&#8722;1) which simplifies to <i>k</i>.<i>k</i>!, as
required.
<p/>
Now we sum the series 1.1!+.....<i>n</i>.<i>n</i>!
<p/>
As we have proved, <i>n</i>.<i>n</i>! is equal to (<i>n</i>+1)! &#8722;<i>n</i>! and therefore (<i>n</i>&#8722;1).(<i>n</i>&#8722;1)! is equal to
(<i>n</i>&#8722;1+1)! &#8722; (<i>n</i>&#8722;1)! which simplifies to <i>n</i>! &#8722;(<i>n</i>&#8722;1)!. If we add the two results, we find that
<i>n</i>! cancels.
<p/>
If we sum the series from 1 to <i>n</i>, we find that
all of the terms cancel except for (<i>n</i>+1)! and
&#8722;(1!). Thus the sum of all numbers of the form
<i>r</i>.<i>r</i>! from 1 to <i>n</i> is equal to (<i>n</i>+1)! &#8722; 1.
<p/>