# 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?

• $\ln(n)-n$ is negative! Do you mean $n-\ln(n)$?
– OmG
Mar 7, 2018 at 21:45
• @OmG, I meant $n ln(n) - n$. Thank you for correction. Mar 7, 2018 at 21:47
• @BarryCipra I have corrected that in text, Too. Thank you for letting me know! Mar 7, 2018 at 21:49

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).

• very nice, I used a similar idea by induction but this is more elegant!
– user
Mar 7, 2018 at 22:14

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 $\frac{n^n}{e^n}\le n!$

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

$$n \log n - n +1 = \int_1^n \; \; \log x \; \; dx < \sum_{j = 2}^n \log j \; = \log n!$$

diagram for $$n=4$$

• Actually $n \log n - n \color{red}{+1} = \int_1^n \; \; \log x \; \; dx$, giving a slightly sharper result. Nov 25, 2019 at 12:23