New year raised to itself

Because $2014 \bmod {9} = 7$ , the sum of digits of $2014^{2014}$ divided by $9$ gives a remainder of $7$ . So, $A \equiv B \equiv C \equiv 7 \pmod {9}$

The number of digits of $2014^{2014}$ is $\lceil \log_{10} 2014^{2014} \rceil = \lceil 2014 \log_{10} 2014 \rceil = 6655$

$\text{max} (A) = 9 \times 6655 = 59895$

$\text{max} (B)$ occurs when $A = 49999$ , $\text{max} (B) = 4 + 9 + 9 + 9 + 9 = 40$

$\text{max} (C)$ occurs when $B = 39$ , $\text{max} (C) = 3+9 = 12$

Because $1 \leq C \leq 12$ and $C \pmod {9} = 7$ , so $C = \boxed{7}$ only.