as reading the book Algorithms by Dasgupta, C.H Papadimitriou. on Page 63.

And it is well known that $\log(n!)≥c\cdot n\cdot\log n$ for some $c > 0$. There are many ways to see this. The easiest is to notice that $n! \ge (n/2)^{(n/2)}$ because $n! = 1 · 2\cdot\cdots\cdot n$ contains at least $n/2$ factors larger than $n/2$; and to then take logs of both sides. Another is to recall Stirling’s formula.

Can someone help to find the proof on $n! \ge (n/2)^{(n/2)}$, or how should we prove it ourselves? As we take log on both side, it became (assuming that its base is some number constant) $\log(n!)$ on the left side, on the right, it became $\log\left((n/2)^{(n/2)}\right)$, further simplify $(n/2)\log(n/2)$ --> $(n/2)\log(n^{-2})$, then finally on the right side: $(-n)\log(n)$. Then we completed the proof (?)

I have two questions, is the proof of $n! \ge (n/2)^{(n/2)}$ corrected? If so, how can we argue back to "$\log(n!)≥c\cdot n\cdot\log n$ for some $c > 0$", if not, how can we proof that "$\log(n!)≥c\cdot n\cdot\log n$ for some $c > 0$"?

  • 1
    $\begingroup$ n! is the product of the numbers 1 to n. The last n/2 of those are all > n/2. $\endgroup$
    – gnasher729
    Aug 13, 2019 at 7:50

2 Answers 2


... further simplify $(n/2)\log(n/2)$ --> $(n/2)\log(n^{-2})$, ...

This is wrong. In fact, for $n\ge 3$, $n^{\log_3 2}\ge 2$, so \begin{align} (n/2)\log(n/2)&=(n/2)(\log n-\log2)\\ &\ge (n/2)\cdot\left(1-\log_3 2\right)\log n\\ &=\left(1-\log_3 2\right)/2\cdot n\cdot \log n, \end{align} and $\log(n!)\ge \left(1-\log_3 2\right)/2\cdot n\cdot \log n$ also holds for $n=1,2$, which completes the proof.


$n! = n \cdot (n-1) \cdot ... \cdot 1 \ge n \cdot (n-1) \cdot ... \cdot (n/2) \ge (n/2)^{(n/2)}$


$\log(n!)≥c\cdot n\cdot\log n$ for $c \ge 1/2$


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