TITLE: proof - Show that $1! +2! +3!+\cdots+n!$ is a perfect power if and only if $n=3$
QUESTION [6 upvotes]: Show that $1! +2! +3!+\cdots+n!$ is a perfect power if and only if $n =3$

For $n=3$, $1!+2!+3!=9=3^2$. I also feel that the word 'power' makes it a whole lot hard to prove. How do we prove this? What technique do we use? 
I don't have any idea how to proceed with this. I would love some hints.

REPLY [3 votes]: Note that,$$1! + 2! + 3! + 4! = 33$$
$$1! + 2! + 3! + 4! + 5! = 153$$
$$1! + 2! + 3! + 4! + 5! + 6! = 873 \ldots $$
The last digit of the numbers is $3$ (This is happening because for $n>4$ the last digit of $n!$ is $0$). Now for a number to be a perfect square the last digit should be one of the digits $1,4,5,6,9$. Hence for $n>3$, $\sum\nolimits_{i = 1}^n {i!} $ cannot be a perfect square.