表明 lg(n！( = Θ(nlgn( 如何证明它？我使用限制来确定顺序，但我卡在某个时候 lim n到 +inf lg(n^n(/lg(n！(
log(n!) = log(n) + log(n-1) + log(n-2) + .... + log(2)
Which can be directly seen as
log(n!) < n*log(n)
log(n!) < c1*n*log(n)
log(n!) > c2*n*log(n)
for n > n0
We have some value of c1 and c2 such that
c1*n*log(n) <= log(n!) <= c2*n*log(n)
Sum ( i =1 to n) i > (n/2)log(n/2) = (n/2)log(n) - (n/2)log(2)
hence we can say that
log(n!) = Big-Theta(nlog(n))