How is log(n!) = Ω( n*log(n))? [closed]

Question

I know that

log (n!) =log (1) + log(2) + .... log(n-1) + log(n) 

and

 n*log(n)= log(n) + log(n) + .... + log(n) or just adding log(n)'s  n times. 

What constant can I multiply n*log(n) that makes it smaller than log(n!)?

I read a few solutions about it being n/2*log(n/2). What constant is that? half?

One solution is from here. Is log(n!) = Θ(n·log(n))?

If C = 1/2, then isn't it just (n/2)*log(n)? How does the n inside of log get affected or why did n become n/2 all of a sudden?

I know the log rule of log(a/b) = log a - log b. Does that rule help?

Answer 1

log(n!) = log(1) + ... + log(n) >= log(n/2) + log(n/2+1) + ... + log(n)
>= n/2*log(n/2) = n/2 *log(n) - n/2*log(2) >= (*) n/2 log(n) - n/4 log(n) 
= n/4 log(n) = 1/4 nlog(n) 

(*) is because for 'high enough' values of n (n>N for some constant N), n/2log(2) < n/4log(n), so we reduce 'bigger' element - which result in lower outcome.

So, we got that log(n!) >= 1/4*nlog(n) - which gives us that it is in Omega(nlogn) by definition with c=1/4.

Regarding the first part, how we got to log(n/2) + log(n/2+1) + ... +log(n)

log(1) + ... + log(n) >= log(1) + ... + log(n) - log(1) - log(2) - ... - log(n/2-1) = 
  = log(n/2) + log(n/2+1) + ... + log(n)