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

Here's a line of attack that leaves you the task of checking only finitely many cases.
The idea is to check divisibility by powers of two. I use the notation $\nu(m)$ to denote
the exponent of the highest power of two that divides a positive integer $m$, so e.g. $\nu(80)=\nu(16\cdot5)=4$, because $2^4=16$.
You have probably seen (or can prove it as an exercise) that
$$
\nu(m!)=\sum_{i=1}^{\lceil \log_2m\rceil} \lfloor\frac m {2^i}\rfloor.\qquad(*)
$$
Also $\nu(mn)=\nu(m)+\nu(n)$, and consequently $\nu(m^n)=n\nu(m)$. Furthermore, if $\nu(a)\neq\nu(b)$ then $\nu(a+b)=\min\{\nu(a),\nu(b)\}$. This latter property is known as non-Archimedean triangle inequality. Let's call it NATI for short.
Here we see that $\nu(2014^{2015})=2015\cdot\nu(2014)=2015$ because $2014$ is even, but not divisibile by four.
Let us first consider the possibility that $a$ is odd. Then so is $a^{2014}$, so $2015\cdot b!$ must be odd. This is possible only, if $b!$ is odd, so we must have $b=0$
or $b=1$ for otherwise $b!$ is even. Check, whether there is matching $a$.
Let us then turn our attention to the possibility that $a$ is even. We first assume that $a$ is not divisible by four or, equivalently, that $\nu(a)=1$. In that case $\nu(a^{2014})=2014$. Because $\nu(2014^{2015})=2015$, NATI restricts the choice of $b$ to those numbers that have $2014=\nu(2015\cdot b!)=\nu(b!)$. Formula $(*)$ comes in handy. I give you the extra bit $\nu(2022!)=2014$. There are very few other possibilities for $b$, because $\nu(b!)$ is a (non-strictly) increasing function of $b$.
Finally, if $\nu(a)>1$, then $\nu(a^{2014})\ge2\cdot2014$, which is greater than $2015$. In this cases NATI forces $\nu(b!)=2015$. This again severely limits the possibilities for $b$.
