Why is $(1 - \frac{1}{p})^n$ close to $e^{-\frac{n}{p}}$ when $n$ and $p$ are large? Looking at this answer by Henry birthday problem - expected number of collisions and struggling to figure out why it matches this other formula provided to me on a programming related question. Thanks!

I don't know if you are familiar with Taylor series, but if yes :
$$\left(1-\frac1p\right)^n=\exp\left(n\ln\left(1-\frac1p\right)\right)=\exp\left(-n\left(\frac1p+o\left(\frac1p\right)\right)\right)$$
and this gives :
$$\left(1-\frac1p\right)^n\sim_{p>>1} \exp\left(-\frac{n}{p}\right)$$
Note than there's no assumption on n.
