If you throw $n$ balls into $n$ bins uniformly and independently at random, let $X$ be the number of balls in the bin with the largest number of balls in it.
Is there a nice elementary way to show $\exists c>0$ such that $\forall n > 1, P(X > c \ln{n}) \leq 1/n$?
This is related to my previous question Expected max load with $n$ balls in $n$ bins? .
There is a reasonably complicated proof of a slightly tighter bound at http://www.ic.unicamp.br/~celio/peer2peer/math/balls-into-bins.pdf .
Let $X_1$ be the number of balls in the first bin. On the event $X_1 \geq k$, one can find a subset of size $k$ of the balls such that all these balls are in the first bin. For a given subset of size $k$, this happens with probability $1/n^k$. So, the union bound gives $$ P(X_1 \geq k) \leq \frac{1}{n^k}\binom{n}{k} $$ Using the other union bound $P(X \geq k) \leq n\times P(X_1 \geq k)$ we find $$ P(X \geq k) \leq n \times \frac{1}{n^k}\binom{n}{k} \leq \frac{n}{k!}. $$
In order to calibrate $k$ such that this is $\leq 1/n$ you can use Stirling's formula or even a simple integral comparison: $$ \log k! = \sum_{i=1}^k \log i \geq k\log k - k -\log k. $$ For example, this is enough to show that for any $n \geq 2$ we have $P(X \geq 4\log n)\leq 1/n$.
