Let $q$ be a polynomial with $deg(q)=2023$ such that : $q(x)=(x+1)p(x)-1.$ Notice that $q$ has $i=\overline{0,2022}$ as its roots, thus we may rewrite $q$ as : $q(x)=x(x-1)\dots(x-2022)A(x),$ where $A$ is a polynomial with $deg(A)=deg(q)-deg(x(x-1)\dots(x-2022))=0.$ Since $q(-1)=-1=(-1)(-1-1)\dots(-1-2022)A(-1)=(-1)^{2023}2023!A(-1),$ we get : $A(-1)=\frac{1}{2023!},$ and $A\equiv \frac{1}{2023!}.$ Finally, $p(2023)=\frac{q(2023)+1}{2024}=\frac{1+1}{2024}=\boxed{\frac{1}{1012}}.$