Let's say that we have a die with $d$ sides. Now we roll the die $n$ times where $n>=d$. What is the probability that we get at least one of each possible results?
I started reasoning like this. The first die does not matter at all, since it will yield a result not taken by any previous dice. The second die have two options. Either it yields the same result as the first, and then the third die will face the same situation as the second die did. The other possibility is that it yields a different result, and in this case the third die will have three different options. But here I got stuck. Well, I could calculate it for a specific case, but would take a very long time of brute forcing when the numbers increase.
I also tried calculating the possibility of NOT getting one of each, but I got stuck there too.
I found out that the probability when throwing exactly $d$ dice is fairly simple. It will be $(d/d)((d-1)/d)$...$(2/d)(1/d)$. So I started to try to find what happens when you increase $n$ by one, but I also got stuck.
Please help. It's not homework. I just started wondering, and it was a couple of years since I did this kind of math.