Determine all possible triplets of integers (n,m,k) satisfying 1!+2!+3!+...+n!=m^k, where n, m and k are greater than 1.

(In reply to some thoughts by Vishal Gupta)

Though we are given that n, m and k are greater than 1, in order for n to satisfy the equation 1!+2!+3!+...+n! = m^k, n must be greater than 3. The factorials for 1, 2 and 3 are already included and the notation in the equation is written that indicates n as some number greater than 3, with m^k equal to the summation of all the factorials of the integers from 1 to n.

Posted by Dej Mar on 2007-02-02 00:24:59