Compute this manually!

$2014 \equiv 5$ ( $\text{mod}$ $7$ ) $\implies 5^{2014} \equiv n$ ( $\text{mod}$ $7$ ).

By Fermat's Little Theorem, $5^{7-1}=5^{6} \equiv 1$ ( $\text{mod}$ $7$ ).

Hence $5^{2014}=(5^{6})^{335} \times 5^{4} \equiv n$ ( $\text{mod}$ $7$ ) $\implies 5^{4} \equiv n$ ( $\text{mod}$ $7$ ).

Since $5^{4}=625 \equiv \boxed{2}$ ( $\text{mod}$ $7$ ), we are done.

$\begin{aligned} 2014^{2014} & \equiv (-2)^{2014} \text{ (mod 7)} & \small \color{#3D99F6} \text{Note that }2014 \equiv -2 \text{ (mod 7)} \\ & \equiv 2^{2014} \text{ (mod 7)} \\ & \equiv 2(2^3)^{671} \text{ (mod 7)} \\ & \equiv 2(8^{671}) \text{ (mod 7)} \\ & \equiv 2(7+1)^{671} \text{ (mod 7)} \\ & \equiv 2(1^{671}) \text{ (mod 7)} \\ & \equiv \boxed{2} \text{ (mod 7)} \end{aligned}$

2014/7 = 287 r.5, 5^n always ends in 5, so 5mod7=2 (idk how to type the equivalence symbol)