JOMO 5, Short 8

It suffices to show that $f(3)$ is going to be so large that it has to be divisible by $2014$ as factorials are super composite. In other words if $f(3)=n!$ and $n\ge2014$ then $2014|n!$ . This means that $f(n)$ for an $n>3$ grows even faster and therefore are all divisible by $2014$ . What's left is $f(1)+f(2)$ which can be computed to be $\boxed{3}$ .

simple......if u take (3!)!)!)...it will clearly be ,more than 2014...so its fact (our function) will be divisible by 2014.....so for numbers more than equal to 3 condition will hold.....remainder will be bcoz of f(1)+F(2)..that is 3