The minimum value of a uniform multinomial distribution Suppose that $n$ balls are randomly put in $k$ boxes, with
uniform chance; called the uniform multinomial
distribution. I'm interested in the chance that no box is
empty. In other words, the chance that the minimum number of balls in any box
is greater than zero.
By the definition of multinomial distribution one obtains:
$$
P(\mbox{no box empty}) = \frac{n!}{k^n}\sum_{\vec{x}} \frac{1}{x_1! \cdots x_k!},
$$
where the sum is over all $k$-vectors $\vec{x}$ with only positive integer entries that sum up to $n$.
I'm wondering whether this formula can be simplified
or approximated.
By the way, this is the "opposite" question of
https://mathoverflow.net/questions/104948 where a formula for
the maximum number of balls in any box is asked.

(Moving my comment to an answer after migration)
The probability that no box is empty, when all boxes are equiprobable, is not difficult to compute by a combinatorial argument. Assuming all the balls and boxes are distinguishable, the number of ways to have no empty box is given by $k!\, S(n,k)$, where $S(n,k)$ is the Stirling number of the second kind. 
Then the desired probability is
$$P=\frac{k!\, S(n,k)}{k^n}$$
An coarse approximation for large values of $n,k$ can be obtained by a "Poissonization" approach: the experiment is asympotically equivalent to throwing $k$ Poisson iid variables with $\lambda = n/k$, and the probability that all greater than zero is
$$P_a = (1-e^{-n/k})^k$$
A few values:
n    k      P              Pa
10   4  0.7806015015    0.7099207499
10   6  0.2718121285    0.2847900333
30  12  0.3591452032    0.3577911635
30   6  0.9748018864    0.9602472297

