Happy 2014th Birthday of the Gregorian Calendar!

$\begin{aligned} \text{Note that:}~~~ \sum_{n=1}^{18} n^{2014} &= &\left(1^{2014}+\cdot\cdot\cdot+8^{2014}+9^{2014}\right)+\left(10^{2014}+\cdot\cdot\cdot+18^{2014}\right) \\ \\ &=&\left(\sum_{n=1}^{9} n^{2014}\right) + \left(\sum_{n=10}^{18} n^{2014}\right) \\ \\ &=&\left(\sum_{n=1}^{9} n^{2014}\right) + \left(\sum_{n=10}^{18} n^{2013}\cdot n\right) \\ \\ &\equiv &\left(\sum_{n=1}^{9} n^{2014}\right) + \left(\sum_{n=1}^{9} \left(-n\right)^{2013}\cdot n\right) \pmod{19} \\ \\ &=& \left(\sum_{n=1}^{9} n^{2014}\right) + \left(\sum_{n=1}^{9} -n^{2014}\right) \\ \\ &=& \left(\sum_{n=1}^{9} n^{2014}\right) - \left(\sum_{n=1}^{9} n^{2014}\right) \\ \\ &=& ~~~ 0 \\ \\ \end{aligned}$

$\text{Therefore:}~~~ \sum_{n=1}^{2014} n^{2014} \equiv 106\cdot \left(\sum_{n=1}^{18} n^{2014}\right) \equiv 106\times 0 = \fbox{0}\pmod{19}$

Python Code:

s=0

for i in range(2014):

a=i+1

for j in range(2014):

    a=(a*(i+1))%19

s=s+a

print s%19