Solve Algebraical.ly $0.5=\dfrac{365!}{365^{n}(365-n)!} $ How does one go about solving this equation? Not sure how to approach this as no factorials will cancel out. Im sorry I meant $\dfrac{365!}{365^{n}(365-n)!}=0.5$. 

We usually solve this equation numerically:
$$a_n=\frac{365!}{365^n(365-n)!}$$
Hence $a_1=1$ and $$a_{n+1}=a_n.\frac{365-n}{365}$$
If you want to solve $a_n=p$, just do a little program that computes $a_n$ from $a_1$ by multiplying at each step by $\frac{365-n}{365}$ until you find $p$.
Here $$a_{23}=0.4927027656$$
