$\left (14^{2014} \right )^{2014} \mod 60$ without a calculator

$$\left (14^{2014} \right )^{2014}=(14^2)^{1007\times2014}\equiv 16^{1007\times2014} \mod 60$$

Next, notice that $16\times 16\equiv 16\pmod{60}$ to conclude $16^n\equiv 16\pmod{60}$ for all integers $n\ge 1$

$$X=\large (14^{2014})^{2014}=14^{2014^2}$$

Now, $60=3\times 4\times 5$. Now,

$$X\equiv\begin{cases}(-1)^{2014^2}\equiv 1\pmod3\\ (-1)^{2014^2}\equiv 1\pmod5\\ 7^{2014^2}\cdot 4^{(2014^2)/2}\equiv 0\pmod4\end{cases}$$

Now, use Chinese Remainder Theorem.

With all the powers of $2$, $(14^{2014})^{2014}$ is clearly equivalent to $0\pmod4$. We also have $(14^{2014})^{2014}\equiv[(-1)^{2014}]^{2014}\equiv 1\pmod{15}$. These $2$ conditions combined yield a result equivalent to $16\pmod{60}$.