Proving that $\sum\limits_{i=1}^k i! \ne n^2$ for any $n$ 
Possible Duplicate:
How to prove that the number 1!+2!+3!+…+n! is never square? 

Show that $\displaystyle\sum\limits_{i=1}^k  i!$ is never a perfect square for $k\ge4$
I could prove $k!$ is never a perfect square using Bertrand's Postulate. But this one seems to be an uphill task.
 A: Your summation looks like $$1!+2!+3!+4!+\cdots+k!=33+5!+\cdots+k!$$Now note that $i!$ is a multiple of $10$ whenever, $i\geq 5$, hence the last digit in your summation is going to be a $3$. By inspection modulo $10$ (I want to look at the decimal place) $$0^2=0$$$$1^2=1$$$$2^2=4$$$$3^2=9$$$$4^2=6$$$$5^2=5$$$$6^2=6$$$$7^2=9$$$$8^2=4$$$$9^2=1$$Hence, it is impossible to have a $3$ in the far most right digit of a square, which your sum does. 
A: Hint $\rm\ mod\ 5\!:\ \ \mathbb Z^2 \equiv \{0,\: \pm 1,\:\pm2\}^2 \equiv \{0,\: \pm 1\} \not\ni -2 \equiv  1!+2!+3!+4! + 5 N$
A: Your last digit at $k=4\;:\; 1!+2!+3!+4!=33$ is $3$ and from $5!$ on you add numbers that end with $0$. So it can never be a square number, since the last digits of the square of any number ends with
$(0^2,(\pm 1)^2,(\pm 2)^2,(\pm 3)^2,(\pm 4)^2,(\pm 5)^2)\mod 10=(0,1,4,9,6,5)$.
A: The following is a brute force approach.  We may calculate the sum at $n=21$, and find  $$\sum_{k=1}^{21}k!=3^2\cdot 11\cdot 877\cdot3203\cdot41051\cdot4699727.$$
Since $11^2$ divides $k!$ for all $k\geq 22$, this implies that $11$ divides the sum exactly once for all $k\geq 21$, and hence the sum is not a square.  For $n\leq 21$, we may check by computer and notice none are squares except $n=3$
