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?

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

3 Answers 3


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).

  • $\begingroup$ very nice, I used a similar idea by induction but this is more elegant! $\endgroup$
    – 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 $n=4$

enter image description here

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

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