find the polynomial such $P(1)P(2)P(3)\cdots P(n)=Q(n!)$ find all polynomial $P(x),Q(x)$ and such all coefficients are real numbers,such that exist infinitely many positive integers $n$ ,such 
$$P(1)P(2)P(3)\cdots P(n)=Q(n!)$$
 A: As the commenters noted, $P(x)=Q(x)=x^k$ and $P(x)=Q(x)=-x^k$ are solutions for any nonnegative integer $k$. We show that these are the only solutions.
Write $P(x) = ax^k + abx^{k-1} + R(x)$ for some real numbers $a,b$ and some polynomial $R(x)$ of degree at most $k-2$ (sorry for the funny normalization of the coefficient of $x^{k-1}$). Then
\begin{align*}
P(1)P(2)\cdots P(n) &= \prod_{x=1}^n \big( ax^k + bx^{k-1} + R(x) \big) \\
&= \prod_{x=1}^n ax^k \prod_{x=1}^n \bigg( 1 + \frac bx \bigg) \prod_{x=1}^n \bigg( 1 + \frac{R(x)}{ax^k(1+b/x)} \bigg).
\end{align*}
(We omit the special case when one of the $1+b/x$ equals $0$.) The first product is simply $a^n(n!)^k$. The logarithm of the second product is
$$
\sum_{x=1}^n \log \bigg( 1 + \frac bx \bigg) = \sum_{x=1}^n \bigg( \frac bx + O\bigg( \frac{b^2}{x^2} \bigg) \bigg) = b\log n + O(1).
$$
In other words, there exist constants $c_1$ and $c_2$ for which
$$
c_1 n^b < \prod_{x=1}^n \bigg( 1 + \frac bx \bigg) < c_2 n^b
$$
for all (sufficiently large) $n$. Similarly, the logarithm of the third product is
$$
\sum_{x=1}^n \log \bigg( 1 + \frac{R(x)}{ax^k(1+b/x)} \bigg) = \sum_{x=1}^n \log \big( 1 + O(x^{-2}) \big) = \sum_{x=1}^n O(x^{-2}) = O(1),
$$
and so the third product itself is bounded between two constants. We conclude that there are constants $c_3$ and $c_4$ such that
$$
c_3 a^n(n!)^k n^b < P(1) P(2) \cdots P(n) < c_4 a^n(n!)^k n^b
$$
for all $n$. (If $a<0$ then this has to be suitably interpreted - the upper and lower bounds alternate with each other.)
On the other hand, if $Q(x) = d x^\ell +{}$lower order terms, then there exist constants $c_5$ and $c_6$ such that
$$
c_5 d(n!)^\ell < Q(n!) < c_6 d(n!)^\ell
$$
for all $n$; in particular,
$$
c_7 \frac{a^n n^b}{d} (n!)^{k-\ell} < \frac{P(1) P(2) \cdots P(n)}{Q(n!)} < c_8 \frac{a^n n^b}{d} (n!)^{k-\ell}
$$
for all $n$. So if the quotient in the middle equals $1$ for infinitely many $n$, we deduce in turn (using the fact that factorials are asymptotically more extreme than exponentials, which are asymptotically more extreme than powers) that $k-\ell=0$ and that $|a|=1$ and that $b=0$. In other words,
$$
P(x) = \pm x^k + R(x) \quad\text{and}\quad Q(x) = dx^k + S(x)
$$
where $S(x)$ has degree at most $k-1$.
(running out of time here, going to quick mode) Consequently, we have
$$
(\pm1)^n \frac{P(1) P(2) \cdots P(n)}{Q(n!)} = \bigg( d + \frac{S(x)}{x^k} \bigg)^{-1} \prod_{x=1}^n \bigg( 1 + \frac{R(x)}{\pm x^k} \bigg).
$$
The right-hand side tends to a limit as $n\to\infty$, and it is eventually monotonic. Therefore, the only way it can equal $1$ infinitely often is if it is identically $1$; this forces $d=1$ and $R(x)=S(x)=0$, which is what we needed to show.
