jmc

The units digit of $1!$ is $1$, the units digit of $2!$ is $2$, the units digit of $3!$ is $6$, the units digit of $4!=24$ is $4$, and the units digit of $5!=120$ is $0$. For all $n\ge 5$, $n!$ is a multiple of $5!$, which is a multiple of 10, so all for all $n\ge 5$, the units digit of $n!$ is 0. This means that the units digit of the sum $1!+2!+3!+4!+5!+\cdots +1000!$ is just the units digit of $1+2+6+4+0+\cdots +0=13$, so the answer is $\boxed{3}$.