We can rewrite the sum as:
\begin{align} \sum_{r = 0}^n r \cdot r! &= \sum_{r = 0}^n (r+1-1) \cdot r! \\ &= \sum_{r = 0}^n [(r+1)!-r!] \end{align} This latter sum telescopes, leaving $$\sum_{r = 0}^n [(r+1)!-r!] = (n+1)! -1.$$
Looking at the results I was curious if there is some way it correlates to arranging $13$ objects in $13$ places without repetition and removing one case. If so, what would be the combinatorial argument for the summation?