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

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?

For $$n\le 365$$, $$p_n=1-\prod_{i=1}^{n-1}\left(1-\frac{i}{365}\right)\ge 1-\prod_{i=1}^{n-1}e^{\frac{i}{365}}=1-e^{\sum_{i=1}^{n-1}\frac{i}{365}}=1-e^{-\frac{n(n-1)}{730}}.$$
• I think a small comment saying $P(collision)=1-P(no collisions)$ will make this a perfect answer. Preemptive (+1) anyway ... – rtybase Feb 11 at 10:38
• How did you get $\sum_{i=1}^{n-1}\frac{i}{365}=\frac{n(n-1)}{730}$? – MinaThuma Feb 11 at 11:17
• @MinaThuma $1+2+3+\ldots + n=n(n+1)/2$. – d.k.o. Feb 11 at 16:48