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). The units digits of $1!+2!+3!+4!$ is the units digit of $1+2+6+4=13$, which is 3. The units digit of the sum $1+2+\dots +9=\frac{9(1+9)}{2}=45$ is 5. Therefore, the units digit of $(1!+1)+(2!+2)+\dots +(9!+9)=(1!+2!+\dots +9!)+(1+2+\dots +9)$ is $3+5=\boxed{8}.$