Proving that $\sum\limits_{i=1}^k i! \ne n^2$ for any $n$ [duplicate]

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.

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.

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 $(0^2,(\pm 1)^2,(\pm 2)^2,(\pm 3)^2,(\pm 4)^2,(\pm 5)^2)\mod 10=(0,1,4,9,6,5)$.

Hint $\rm\ mod\ 5\!:\ \ \mathbb Z^2 \equiv \{0,\: \pm 1,\:\pm2\}^2 \equiv \{0,\: \pm 1\} \not\ni -2 \equiv 1!+2!+3!+4! + 5 N$