Consider a balls-and-bins game with m balls and n bins. Each ball is independently thrown into a bin that is chosen uniformly at random.
Question: Show that, for some $k$, the probability that at least one bin gets more than $k$ balls is bounded by $$\left(\frac{m}{n}\right)^k\frac{n}{k!}$$
Approach:
I already know that if $E_i$ is the event that the $i$ th bin contains more than or equal to $k$ balls, then $$ p\left(E_i\right) \leq\begin{array}{c} m \choose k \end{array}\left(\frac{1}{n}\right)^k $$ and then we take union bound over all bins and by Stirling's approximation, the total probability is bounded on the above by $$\left(\frac{m}{n}\right)^k \frac{n}{k !} $$ But here in the question we have strictly more than $k$ balls and yet they ask to prove the same bound. So I am wondering if there is something wrong with the question or can my solution be modified slightly to prove the required claim.