Probability of Selecting a Server This problem is related to randomly assigning servers to user and it is generalized based on $N$ users and $K$ servers.

Each of $N$ users requesting a certain website is assigned to a given
  server out of $K$ servers. Suppose that this assignment is done for each
  user independently by selecting one of the $K$ servers uniformly at
  random. The question is: What is the probability that there are no
  servers that are no assigned any users?

I wasn't sure how to approach this problem. I was trying to do something like 
$$P[\text{all users have a server}] = \prod_{i=0}^{K-1} \frac{(K - i)}{K}$$ This would be the probability of picking a different server. 
 A: There is unfortunately no closed form for the probability, unless you consider expressions in terms of Stirling Numbers of the Second Kind closed form. 
There are $K^N$ equally likely functions from the set of users to the set of servers. To count the favourables, we count the number of onto (surjective) functions. The simplest explicit expression is obtained by Inclusion/Exclusion. The number of surjective functions is
$$K^N -\binom{K}{1}(K-1)^N +\binom{K}{2}(K-2)^N-\binom{K}{3}(K-3)^N+\cdots.$$
A: I was curious about Andre's answer and I looked at particular example. Namely $K=3, N=5.$
First of all we set servers and users in order and we get that there are $3^5$ options.
So there are two situations where we have no assigend servers. 
$(\star)$ One server is without any user. Choice of the server gets us $\binom{3}{1}$ and all users goes to the other two ones. Hence we get $(3-1)^5.$ But we have to exclude the situation when all goes to one server. There are two such situations, but write it $\binom{2}{1}(3-2)^5.$ 
$(\Delta)$ Two servers are without any user. Choice of two servers without users gives us $\binom{3}{2}$ and all users goes to last one. Hence $(3-2)^5=1^5.$
Your question is about situation where no servers are free of users. Hence the resut is just complement of these two situations above. Namely
$$3^5-\overbrace{\left[\binom{3}{1}(3-1)^5-\binom{2}{1}(3-2)^5\right]}^{\star}-\overbrace{\left[\binom{3}{2}(3-2)^5\right]}^\Delta=3^5-\binom{3}{1}(3-1)^5+\binom{3}{2}(3-2)^5.$$ As we are interested in probability we have to divide it by $3^5.$
It looks like Andre's answer, but using my way of reasoning, it is not easy (for me) to guess the general formula. Maybe doing larger case line $K=4,N=6$ would give more apparent result.
