then there exist $m,n$ such $(m!)\cdot (n!)\equiv r\pmod p$ let $p$is give odd prime number,show that: for any postive integer $r$,then there exists postive ineteger $m,n$ such that
$$(m!)\cdot (n!)\equiv r\pmod p$$
I try use this $$(m)!\equiv (p+1)(p+2)\cdots (p+m) \pmod p$$
$$(n)!\equiv (p+1)(p+2)\cdots (p+n) \pmod p$$
so
$$(m!)\cdot (n!)\equiv (p+1)(p+2)\cdots (p+m) (p+1)(p+2)\cdots (p+n)\pmod p$$
But I can't it
I am a tenth grader, so I find this a somewhat hard problem. Hence, I'd be doing it the other way around.
If $\exists x : x \leq m , p \mid x$ or $\exists x : x\leq n, p \mid x \implies r = 0 \implies m,n &lt; p$ for $r \neq 0$ 
The only easiest thing we can do is to prove that $\exists r :m! \times n! \equiv r (\mod{p})$ \
Since $m &lt; p, m \equiv m \pmod{p}$.
If $m! &gt; p$, let $m! = px + a \implies ,m! \equiv a \pmod{p} (\implies a = m! - px)$
Also, if $n &lt; p$, $n \equiv n \pmod{p}$.
If $n! &gt; p$, let $n! = py + b \implies n! \equiv \pmod{p}(\implies b = n! - px)$
If we take $r = a\times b$, $r = (m! - px)(n! - py) = p(-m!y -n!x + xy) + m!n! = p(-pxy -ay -pxy -bx + xy) +m!n! = p(xy(1- 2p) -(ax + by)) + m!n! $ 
Since the term before the plus sign is divisible by $p$, we can say $r \equiv m!n! \pmod{p}$
Or, if you take $m!n!$, you get $p^2xy + pxb + pya + ab = pq + r (r = ab)$. Thus, $m!n! \equiv ab \pmod{p} \implies m!n! \equiv r \pmod{p}$
This necessarily means that for most $r$'s there exists some $m,n$ satisfying $m!n! \equiv r \pmod{p}$. Why I say most $r$'s is because there could be exceptions (as JMP pointed out in the comments). Also, you can't all of a sudden generalise this. As in the comments, simple algebra won't work here.
This solution might be faulty.. please notify me if it is so.
I am a tenth grader, so I find this a somewhat hard problem. Hence, I'd be doing it the other way around.
If $\exists x : x \leq m , p \mid x$ or $\exists x : x\leq n, p \mid x \implies r = 0 \implies m,n &lt; p$ for $r \neq 0$ 
The only easiest thing we can do is to prove that $\exists r :m! \times n! \equiv r (\mod{p})$ \
Since $m &lt; p, m \equiv m \pmod{p}$.
If $m! &gt; p$, let $m! = px + a \implies ,m! \equiv a \pmod{p} (\implies a = m! - px)$
Also, if $n &lt; p$, $n \equiv n \pmod{p}$.
If $n! &gt; p$, let $n! = py + b \implies n! \equiv \pmod{p}(\implies b = n! - px)$
If we take $r = a\times b$, $r = (m! - px)(n! - py) = p(-m!y -n!x + xy) + m!n! = p(-pxy -ay -pxy -bx + xy) +m!n! = p(xy(1- 2p) -(ax + by)) + m!n! $ 
Since the term before the plus sign is divisible by $p$, we can say $r \equiv m!n! \pmod{p}$
Or, if you take $m!n!$, you get $p^2xy + pxb + pya + ab = pq + r (r = ab)$. Thus, $m!n! \equiv ab \pmod{p} \implies m!n! \equiv r \pmod{p}$
This necessarily means that for most $r$'s there exists some $m,n$ satisfying $m!n! \equiv r \pmod{p}$. Why I say most $r$'s is because there could be exceptions (as JMP pointed out in the comments). Also, you can't all of a sudden generalise this. As in the comments, simple algebra won't work here.
This solution might be faulty.. please notify me if it is so.