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?

Hint: $n!\equiv 0 \pmod{5}$ for all $n\ge 5$.
