A professor in class gave the following lower bound for the factorial $$ n! \ge {\left(\frac n2\right)}^{\frac n2} $$ but I don't know how he came up with this formula. The upper bound of $n^n$ was quite easy to understand. It makes sense. Can anyone explain why the formula above is the lower bound?

Any help is appreciated.

  • $\begingroup$ Can you prove it by induction? $\endgroup$
    – Ian Coley
    Nov 21, 2013 at 0:24

2 Answers 2


Suppose first that $n$ is even, say $n=2m$. Then

$$n!=\underbrace{(2m)(2m-1)\ldots(m+1)}_{m\text{ factors}}m!\ge(2m)(2m-1)\ldots(m+1)>m^m=\left(\frac{n}2\right)^{n/2}\;.$$

Now suppose that $n=2m+1$. Then

$$n!=\underbrace{(2m+1)(2m)\ldots(m+1)}_{m+1\text{ factors}}m!\ge(m+1)^{m+1}>\left(\frac{n}2\right)^{n/2}\;.$$


Hint: For a positive integers $a$, we have $a(n-a)\ge \frac{n}{2}$. This is because for $0\le x\le n$, the function $x(n-x)$ is increasing up to $x=\frac{n}{2}$, and then decreasing.

Pair the numbers $a$ and $n-a$ in the factorial.

  • 1
    $\begingroup$ Can you please prove the inequality in the hint? I don't see how $x(n-x)$ having a maxima at $x=n/2$ explain the inequality. $\endgroup$ May 7, 2020 at 22:28
  • $\begingroup$ @ShivTavker You can assume x is a real number and diferentiate the equation and equal it to 0: -x^2 +xn = 0 -> -2x + n = 0 -> x = n / 2. $\endgroup$ Apr 16, 2021 at 4:41

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