on our lesson at our university, our professsor told that factorial has these estimates
$n^{\frac{n}{2}} \le n! \le \left(\dfrac{n+1}{2}\right)^{n}$
and during proof he did this
$(n!)^{2}=\underbrace{n\cdot(n-1)\dotsm 2\cdot 1}_{n!} \cdot \underbrace{n\cdot(n-1) \dotsm 2\cdot 1}_{n!}$
and then:
$(1 \cdot n) \cdot (2 \cdot (n-1)) \dotsm ((n-1) \cdot 2) \cdot (n \cdot 1)$
and it is equal to this
$(n+1)(n+1) \dotsm (n+1)$
why it is equal, I didn't catch it. Do you have any idea? :)
Not equal, but by a standard inequality: $\sqrt{ab} \le \frac{a+b}{2}$, so $ab \le \frac{(a+b)^2}{4}$.
So all products $i\cdot ((n+1)-i)$ are estimated above by $\frac{(i + ((n+1)-i))^2}{4} = \frac{(n+1)^2}{4}$. So no equality, but upper bounded by. Because we have $(n!)^2$ we get rid of the square roots again so $n! \le (\frac{n+1}{2})^n$ (where the $n$-th power comes form the fact that we have $n$ terms in the $(n!)^2$ expression.
The first equality is just rearranging terms.
$$n^{n/2}\le n!\le \left(\frac{n+1}{2}\right)^{n}$$
$$\iff n^n\le (n!)^2\le \left(\frac{(n+1)^2}{4}\right)^n$$
Now, $n\le i((n+1)-i)$ for all $i\in\{1,2,\ldots,n\}$, because this is equivalent to $n(1-i)\le i(1-i)$. If $i=1$, then it's true. If $i\neq 1$, then this is equivalent to $n\ge i$, which is true.
Also $ab\le \frac{(a+b)^2}{4}$ for all $a,b\in\mathbb R$, because this is equivalent to $(a-b)^2\ge 0$, which is true.
$$n^n=\prod_{i=1}^n n\le \underbrace{\prod_{i=1}^n i((n+1)-i)}_{(n!)^2}\le \prod_{i=1}^n\frac{(i+((n+1)-i))^2}{4}=\left(\frac{(n+1)^2}{4}\right)^n$$
