TITLE: Prove that $n \ln(n) - n \le \ln(n!)$ without Stirling
QUESTION [5 upvotes]: I need to prove that  $n \ln(n) - n \le \ln(n!)$. I have solved this but I've used the Stirling substitution for the factorial term which does not seem good to me in this proof. I am sure that there must be a direct way to solve this. 
One way I can think about tackling this problem is simply breaking the left and right hand side into primary terms: 
$$\ln(n!) = \ln(n) + \ln(n-1) + \ln(n-2) + \dots + \ln(2) + \ln(1)$$
$$n \ln(n) - n = n (\ln(n)-1) = (\ln(n) - 1) + (\ln(n) -1) + \dots + (\ln(n) -1)$$
 
I need to somehow show that the the top expression is greater than the bottom expression, but I can only be sure that $\ln(n) > \ln(n) -1$ and that $ \ln(n-1)>\ln(n) - 1$ 
What can I do about the rest?

REPLY [9 votes]: As an alternative note that
$$ n \ln n - n \le \ln(n!)\iff e^{n \ln n - n}\le e^{\ln(n!)}\iff \frac{n^n}{e^n}\le n!$$
which can be proved by induction as follow
base case

$n=1 \implies  \frac1e \le 1$

induction step

assume $\frac{n^n}{e^n}\le n!$
$\frac{(n+1)^{n+1}}{e^{n+1}}=\frac{n+1}{e}\frac{(n+1)^n}{n^n}\frac{n^n}{e^n}\le \frac{n+1}{e}\left(1+\frac1n\right)^n n!\le(n+1)n!=(n+1)!$