Prove that $n \ln(n) - n \le \ln(n!)$ without Stirling 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?
 A: 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 $\frac{n^n}{e^n}\le n!$

*$\frac{(n+1)^{n+1}}{e^{n+1}}=\frac{n+1}{e}\frac{(n+1)^n}{n^n}\frac{n^n}{e^n}\le \frac{n+1}{e}\left(1+\frac1n\right)^n n!\le(n+1)n!=(n+1)!$
A: $$ n \log n - n +1 = \int_1^n \; \; \log x \; \; dx < \sum_{j = 2}^n \log j \; = \log n!  $$
diagram for $n=4$

A: Recall that the sequence 
$$ e_n = \left( 1+ \frac 1n \right)^n $$
is increasing and converges to $e$. Thus, 
$$ e^n \ge e_1 \cdot e_2 \cdot \ldots \cdot e_n = \frac{(n+1)^n}{n!}. $$
In particular we obtain the weaker inequality $e^n \ge \frac{n^n}{n!}$, which is equivalent to $n \ln n - n \le \ln (n!)$. 

A by-product of this is that the sequence 
$$ \sqrt[n]{e_1 \cdot e_2 \cdot \ldots \cdot e_n} = \frac{n+1}{\sqrt[n]{n!}} $$
is increasing and tends to $e$ (by an application of Stolz-Cesaro theorem). 
