Show that if $n\geq 5$ then $1!+2!+3!+\ldots+n!\equiv 3 \pmod{5}$ Show that if $n\geq 5$ then $1!+2!+3!+\ldots+n!\equiv 3\pmod{5}$
I found this problem  in Legendre Symbol Section, but I can't find relationship between $n!$ between factorial and Legendre Symbol.
Anyone can give a hint or some part of proof?

We have that for $n\geq 5$ then 
$$1!+2!+3!+4!+5!+\dots +n!=33+5\cdot(4!+\dots +(n!/5))\equiv 33\equiv 3\pmod{5}.$$
Note that for $n\geq 5$, the number $(4!+\dots +(n!/5))$ is an integer because $5$ divides $k!$ for $5\leq k\leq n$.
