Integer solutions to $a^{2014} +2015\cdot b! = 2014^{2015}$ 
How many  solutions are there for $a^{2014} +2015\cdot b! = 2014^{2015}$, with $a,b$ positive integers?

This is another contest problem that I got from my friend.
Can anybody  help me  find the answer? Or  give me a hint to solve this problem?
Thanks

Not the whole solution, but enough to get going.
We have that $\displaystyle a^{2014}+b!\equiv 0 \pmod{2014}$


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*If $b\geq 53$, then $\displaystyle a^{2014}\equiv 0 \pmod{2014}$. 


Since $2014=2\times19\times53$, this implies $\displaystyle a\equiv 0 \pmod{2014}$
Hence $a=2014k$ for some positive $k$
Plugging this in yields $\displaystyle 2014^{2014}k^{2014}+2015  b!=2014^{2015}$
Hence $2014^{2014}(2014-k^{2014})=2015  b!$
The RHS being positive, so is the LHS, and thefore $2014\geq k^{2014}$, which forces $k=1$
Therefore $a=2014$. Furthermore, $2014^{2014}(2013)=2015  b!$ is clearly a contradiction.


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*Thus, $b\leq 52$

