prove Diophantine equation has no solution $\prod_{i=1}^{2014}(x+i)=\prod_{i=1}^{4028}(y+i)$ 
show that this equation
  $$(x+1)(x+2)(x+3)\cdots(x+2014)=(y+1)(y+2)(y+3)\cdots(y+4028)$$
  have no positive integer solution.

This problem is china TST (2014),I remember a famous result? maybe is a Erdos have proof this problem? 
maybe this follow have some paper?
$$(x+1)(x+2)(x+3)\cdots(x+k)=(y+1)(y+2)(y+3)\cdots(y+2k)$$
positive integer solution?
Thank you for help 

This is not an answer, but it's too long for a comment. Perhaps it helps:
Observe that $$\frac{(y+1)\cdots(y+4028)}{4028!}={y+4028\choose y},$$ which is an integer. Then $$(x+1)\cdots(x+2014)=4028!{y+4028\choose y},$$ so the left hand side is an integer multiple of $4028!$.
This yields a large lower bound for $x$ as follows. The product $(x+1)\cdots(x+2014)$ must be divisible by the product of $p$, $2p$, and $3p$, for any prime with $3p\le4028$. If $p>1007$, there isn’t room for three multiples of $p$ between $x+1$ and $x+2014$, so one of the $x+j$ must be a multiple of $p^2$. Using $p=1327$, then $x>1327^2$.
You can also get some partial congruences for $x$. The prime $4027$ divides $(x+1)\cdots(x+2014)$, so $(x \mbox{ mod } 4027)\ge2013$. 
[Note that in my comment to the original question, I provide a reference to a general result implying that there is no solution to the original equation.]
