If $N$ is a $n$-bit number, how many bits longs is $N!$, approximately in $\Theta( )$ form?
I know that $$ \log(N!) = \log(N*(N-1)*...*2*1) \leq \log (N)+\log (N-1)+...+\log(2)+\log(1) $$ $$ \log(N!) = O(N\log(N)) $$ I'm unsure about how to show it in terms of $\Theta$ instead of O notation
