The sum $1!+2!+3!+...+2007!$ is not a perfect square

Question

Today my teacher told me to prove this:- Prove that $1!+2!+3!+...+2007!=\sum_{n=1}^{2007}(n!) $ is neither a perfect square nor a perfect cube. Not getting any idea. Please help. Thanks in advance.

Answer 1

All factorials starting with $5!$ are multiples of $10$ and thus have ones digit $0$. Since $1! + 2! + 3! + 4! = 33$, the number $\displaystyle \sum_{n=1}^{2007} n!$ has ones digit $3$. Can it be a perfect square?

Answer 2

$n!$ is a multiple of $7$ for $n\ge 7$. Hence the sum is $\equiv 1!+2!+3!+4!+5!+6!\equiv 5\pmod 7$. Hence if your sum is a square $a^2$, then $a^6\equiv 5^3\equiv 6\pmod 7$, and if it is a cube $a^3$, then $a^6\equiv 5^2\equiv 4\pmod 7$. But $a^6\equiv 1\pmod 7$ by Fermat if $7\not\mid a$ , and $a^6\equiv 0\pmod 7$ if $7\mid a$. Therefore, the sum is neither a square nor a cube.

Answer 3

Hint: What is your sum modulo $7$? What are the possible squares and cubes modulo $7$?