Show that $p_{n} \geq 1- \exp{(-n(n-1)/730)}$

Question

On the issue of the birthday paradox,Let $p_{n}$ be the probability that in a class of $n$ at least $2$ have a their birthdays on the same day (exclude $29$ Feb). Use the inequality $1-x \leq e^{-x}$ to show that:

$p_{n} \geq 1- \exp{(-n(n-1)/730)}$ and then determine $n \in \mathbb N$ so that $p_{n} \geq \frac{1}{2}$

My ideas:

First $p_{n}=1-\frac{\frac{365!}{(365-n)!}}{365^{n}}$ using the inequality given to us.

$1-\exp{(-\frac{\frac{365!}{(365-n)!}}{365^{n}})}\geq p_{n}$, what am I supposed to do next? Use Stirling's Formula?

Answer 1

For $n\le 365$, $$ p_n=1-\prod_{i=1}^{n-1}\left(1-\frac{i}{365}\right)\ge 1-\prod_{i=1}^{n-1}e^{\frac{i}{365}}=1-e^{\sum_{i=1}^{n-1}\frac{i}{365}}=1-e^{-\frac{n(n-1)}{730}}. $$