How to prove n! is equivalent to $n^n$?

Question

I have been using $n!$ as $n^n$ when talking about big-$O$ and time complexity but never exactly knew why? I searched also but couldn't find anything. Somebody, please help me understand.

Answer 1

$n!$ does not grow as fast as $n^n$ as $n\to \infty.$

Method 1:For integer $m\geq 1$ we have $\log m<\int_m^{m+1}\log x\;dx$ so for integer $n\geq 1$ we have $$\log (n!)=\sum_{m=1}^n\log m<\sum_{m=1}^n\int_m^{m+1}\log x \;dx=\int_1^{n+1}\log x \;dx=$$ $$=(n+1)\log (n+1) -n.$$ Subtracting $n\log n$ from this and re-arranging, we have $$\log (n!)-\log (n^n)<\log ((1+1/n)^n)-(n-\log (n+1))$$ and the RHS above goes to $-\infty$ as $n\to \infty.$

Method 2. (Simpler). For odd $n$ let $n'=(n+1)/2.$ For even $n$ let $n'=n/2.$

For $n\geq 2$ we have $$n!=(\prod_{j=1}^{n'}j)\cdot (\prod_{j=n'+1}^nj)\leq (n')^{n'}\cdot n^{n-n'}.$$ So for $n\geq 2$ we have $$n!/n^n\leq (n')^{n'}\cdot n^{n-n'}/n^n=(n'/n)^{n'}\leq \left(\frac {n+1}{2n}\right)^{n'}\leq \left(\frac {n+1}{2n}\right)^{(n/2)}\leq (3/4)^{(n/2)}$$

Answer 2

A crude but easy answer is that to compute $n!$, you are multiplying $n$ factors which are, on average, of order $n$.

More precisely, let's estimate $\log (n!)$ instead of $n!$. We have $\log n!=\log 1+\log2+\log3+\dotsb+\log n$. This expression looks like a Riemann sum for $\int\log x\,dx$. It could be the right-hand Riemann sum for $\int^n_0\log x\,dx$, which gives us the right upper bound, but is a divergent integral. Or it could be the left-hand Riemann sum for $\int^{n+1}_1\log x\,dx$, which is convergent, but has not the right upper bound we want (but would give a valid approximation for $n!$). Let's instead use the trapezoidal sum:

$$ \frac{1}{2}\log 1 + \log 2 + \log 3 + \dotsb + \log(n-1) + \frac{1}{2}\log n \approx \int^n_1\log x\,dx $$ which has the right upper bound, and a lower bound that won't be divergent.

On the left-hand side of this approximation, we have $$\frac{1}{2}\log 1 + \log 2 + \log 3 + \dotsb + \log(n-1) + \frac{1}{2}\log n \\ = \log 1 + \log 2 + \log 3 + \dotsb + \log(n-1) + \log n - \left(\frac{1}{2}\log 1 + \frac{1}{2}\log n\right)\\ = \log(n!) - \left(\frac{1}{2}\log 1 + \frac{1}{2}\log n\right) $$

And on the right-hand side, we have by using integration by parts

$$ \int^n_1\log x\,dx = (x\log x-x)\Big|^n_1 = n\log n -n + 1 $$

Putting it together, we have

$$ \log(n!) \approx\left(\frac{1}{2}\log 1 + \frac{1}{2}\log n\right) + n\log n -n + 1\\ =(n+\frac{1}{2})\log n-n+1. $$

Exponentiation gives

$$ n! \approx \frac{en^{n+\frac{1}{2}}}{e^n}=e\sqrt{n}\left(\frac{n}{e}\right)^n. $$

So keeping only the leading order terms in each factor, we have that $\log(n!)$ is asymptotic to $n\log(n)$, which gives you the crude estimate $n!\approx n^n$ that you were looking for.

Note that dropping sub-leading order terms yields an asymptotic expression for the logarithm, but this is no longer true after exponentiating, so the expression $n!\approx n^n$ should not be taken literally.

Answer 3

Actualy $n!=o(n^n)$

Here is a simple way to see this:

Let $a_n=\frac{n!}{n^n}$ then:

$$\frac{a_{n+1}}{a_n}=\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}}=\frac{n^n(n+1)!}{n!(n+1)^{n+1}}=\frac{n^n}{(n+1)^n}=(\frac{n}{n+1})^n \to \frac{1}{e}<1$$

Thus from ratio test for sequences we have that $a_n \to 0$