log(N!) = log(1) + log(2) + ... + log(N)
       

If we keep only the last N/2 terms

 log(N!) > log(N/2) + log(N/2 + 1) + ... + log(N)
   

Now replace all the terms by log(N/2) and the sum is even smaller:

 log(N!) > log(N/2) + log(N/2) + ... + log(N/2) = 1/2 * Nlog(N/2)
   

Since log(N/2) = log(N) - 1,

      log(N!) > 1/2 * Nlog(N) -