Solution:

We claim $$\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+\dots+\dfrac{n-1}{n!}$$$$= 1- \dfrac 1{n!}.$$

To prove this, we can use induction.

If $n=2,$ then the sum is $\dfrac 1{2!} = 1-\dfrac 1{2!} .$

If it works for $n-1,$ then $$\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+\dots+\dfrac{n-1}{n!}$$ $$= 1- \dfrac 1{(n-1)!}+\dfrac{n-1}{n!}$$$$= 1- \dfrac n{n!}+\dfrac{n-1}{n!} = 1 - \dfrac 1{n!}.$$

This means our formula is proven, so we can get our answer by plugging in $n=2022.$ Our answer is $1- \dfrac 1{2022!},$ yielding $a=1,b=2022,$ making our answer $2023.$

Thus, our answer is D .