1!+2!+3!=9=3^{2}.
Prove that k=3 is the only case of the sum 1!+2!+3!+...k!
resulting in an integer power of an integer number.
(In reply to
The easy half by Jer)
My previous comment is basically dealing with mod 10 to rule out squares.
I haven't found a method that works for all prime powers, but with different moduli one can rule out different powers.
For example, using mod 7, it's easy to rule out cubes.

Posted by Jer
on 20150225 10:43:51 