The number of bits in $N!$ I'm struggling with this homework problem:
If $N$ is an $n$-bit number, how many bits long is $N!$ approximately (in $Θ(·)$ form)?
I know that the number of bits in $N$ is equivalent to log base $2$ of $N$, but that's about it. Any hints on where to start?

$$\log_2(n!)=\log_2[n\cdot (n-1)\cdots 1]=\log_2 n+\log_2(n-1)+\cdots+\log_2 1$$
There are $n$ terms, and most of the terms are close to $\log_2 n$ in magnitude. Therefore $\log_2(n!)=\Theta(n\log n)$.
(To prove this more formally, use Stirling's approximation.)
