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 $\dfrac{365!}{365^{n}(365-n)!}=0.5$. 


/* birthProb_0.cc  25-nov-2013
http://math.stackexchange.com/users/85343/felix-marin

http://math.stackexchange.com/questions/580287/solve-algebraical-ly-0-5-dfrac365365n365-n/580318#580318
*/
#include <cmath>
#include <iostream>
using namespace std;
typedef long double ldouble;
typedef unsigned long long ullong;
const ldouble LN365=log(365.0L);

inline ldouble lnFac(ullong n)
{
 return lgammal(static_cast<ldouble>(n) + 1.0L);
}
const ldouble LN365FAC=lnFac(365ULL);

inline ldouble f(ullong n)
{
 return exp(LN365FAC - n*LN365 - lnFac(365ULL - n));
}

int main()
{
 ullong n=0;
 while (   ( n<=365 ) && ( f(n)>0.5L )  ) ++n;

 (--n),cout<<"n = "<<n<<" ---> "<<f(n)<<endl;
 (++n),cout<<"n = "<<n<<" ---> "<<f(n)<<endl;
 (++n),cout<<"n = "<<n<<" ---> "<<f(n)<<endl;

 return 0;
}

Result:
$$
\begin{array}{l}
{\tt n = 22 ---> 0.524305}
\\
\color{#0000ff}{\large{\tt n = 23 ---> 0.492703}}
\\
{\tt n = 24 ---> 0.461656}
\end{array}
$$
