How do you calculate probability of rolling all faces of a die after n number of rolls? [duplicate]

The probability of not rolling a 1 in $n$ rolls is $(5/6)^n$, similarly for not rolling a $2,\ldots,6$. Now, $6(5/6)^n$ would be the probability that we are not rolling a $1,\ldots,6$, but we would be double counting the rolls where we do not roll both a 1 or a 2. The probability of not rolling two specified numbers in $n$ rolls is $(4/6)^n$ and there are $\binom{6}{2}$ pairs of numbers. But if we subtract these out we undercount the rolls that avoid three numbers. This generalizes to the inclusion-exclusion principle, giving us the probability of missing any number as $$\binom{6}{1}(5/6)^n-\binom{6}{2}(4/6)^n+\binom{6}{3}(3/6)^n-\binom{6}{4}(2/6)^n+\binom{6}{5}(1/6)^n$$ as the probability of missing at least one number in $n$ rolls. The probability of rolling all of them is just 1 minus this probability.