n=1 k = 1! = 1, a perfect square
n=2 k = 1! + 2! = 1 + 2 = 3
n=3 k = 1! + 2! + 3! = 1 + 2 + 6 = 9, a perfect square
n=4 k = 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33
n=5 k = 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 + 120 = 153
... ... ... ... ... ... ...
Since 5!, which is 120, ends with a 0, the factorial of every
integer 5 or higher will also end in a 0 because it will be a
multiple of 120. Since 1! + 2! + 3! + 4! = 33 ends with 3,
the last digit of every sum after that will also end with a 3,
That's because we will only be adding to it integers that end
in 0. Since no perfect square can end with 3, 1! + 2! + 3! = 9
is the last sum above that will be a perfect square.
So only the values n=1 and n=3 produce a sum for which k is
a perfect square. So the answer is 2 values of n.
Edwin
