jmc

For $n\ge 5$, $n!$ includes the product $2\times 5=10$, which means the units digit of $n!$ is 0 (since 0 multiplied by any number is 0). Therefore, it only remains to find the units digit of $1!+2!+3!+4!$, which is the units digit of $1+2+6+4=13$. Thus, the units digit of $1!+2!+\dots +50!$ is $\boxed{3}$.