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