Prove: There exist real numbers a,b, such that 1<a<b and for any integer $n\ge 2$, there exist positive integers $x\_1\ge x\_2\ge ...\ge x\_{\lfloor{\frac{n}{2}}\rfloor}$such that $n!=\Pi\_{k=1}^{\lfloor\frac{n}{2}\rfloor}x\_k$ and $x\_k\in (a^\frac{n}{k},b^{\frac{n}{k}})$.


I've tried to list all the prime factors of n! but that's too complicated. Notice that the function $x^\frac{n}{k}$ is too strange. It means there are many small factors and a few large factors.



Factorising n! fitting $x^\frac{n}{k}$
Your initial guess about prime factors can produce a solution. More precisely, it is known that
$$
n!=\prod\_{p\leq n} p^{v\_p(n!)},
$$
where the product is taken over prime numbers and
$$
v\_p(n!)=\left[\frac{n}{p}\right]+\left[\frac{n}{p^2}\right]+\ldots
$$
Let now for any $1\leq k\leq n/2$
$$
x\_k=\prod\_{\alpha, p: p^\alpha\leq n/k}p.
$$
The product is over prime values of $p$ and positive integers $\alpha$. The inequality $x\_{k}\geq x\_{k+1}$ is trivial. Product of all $x\_k$ is equal to $n!$ by the factorisation above: the factor $p$ that corresponds to a given $d=p^\alpha$ appears in $x\_k$ for $d\leq n/k$, there are $\left[\frac{n}{p^\alpha}\right]$ such $k$. This means that
$$
\prod\_k x\_k=\prod\_{\alpha,p} p^{\left[\frac{n}{p^\alpha}\right]}=\prod\_{p\leq n} p^{v\_p(n!)}=n!,
$$
as needed. Finally, let $\Lambda(d)$ denote the von Mangoldt function, i.e. $\Lambda(d)=\ln p$ if $d=p^\alpha$ for some $p$ and $\alpha$ and $\Lambda(d)=0$ otherwise. Then we clearly have
$$
\ln x\_k=\sum\_{d\leq n/k} \Lambda(d)=\psi(n/k).
$$
Here $\psi(x)$ is Chebyshev's psi function, the summatory function of $\Lambda$. It is well-known that there are $c,C>0$ such that for $x\geq 2$ we have
$$
cx\leq \psi(x)\leq Cx.
$$
In particular, one can take $1<a<e^c$, $b>e^C$ and get the desired bound:
$$
a^{n/k}\leq \exp(\psi(n/k))=x\_k\leq b^{n/k}.
$$


