We use a stars-and-bars approach, introducing five partitions that divide the $n$ stars (dice rolls) into six parts, some of which may be empty. When the partitions are interpreted as rolls of 1, 2 and so on, the roll sequence is weakly increasing, and all such roll sequences are in bijection with all partitions of the $n$ stars. We have $\binom{n+5}5$ possible partitions, and thus roll sequences satisfying the weakly increasing condition. The final probability is $\frac{\binom{n+5}5}{6^n}$.