Show that the probability of $m$ balls being randomly selected with replacement are all different is approximately $e^{-m^2/2n}$ Show that the probability of m balls being randomly selected with replacement are all different is approximately  $e^{-m^2/2n}$.
I've been fiddling around with equations involving derangements in order to get the $e$, but I have no idea where the $-m^2$ fits into the equation at all. Can anyone help with this?

For $m \ll n$ we have
$$\begin{align}
\mathsf P(m \text{ balls are different}\mid m \text{ balls selected})
        & = \prod_{k=1}^{m-1}\left(1-\frac{k}{n}\right)
\\[1ex] & = \exp\left(\sum_{k=1}^{m-1}\ln\left(1-\frac{k}{n}\right)\right)
\\[1ex] & \approx\exp\left(-\sum_{k=1}^{m-1}\frac{k}{n}\right)
\\[1ex] & = \exp\left(-\frac{1}{2}\frac{m(m-1)}{n}\right)
\end{align}$$
