To get the unit digit of $2017^{2017}$, we will find $2017^{2017}\;mod\;10.$
$2017^{2017}\;mod\;10$$=2017^{2016+1}\;mod\;10$
$=(2017^{2016}*2017^{1})\;mod\;10$
$=(2017^{2016}\;mod\;10*2017^{1}\;mod\;10)\;mod\;10$
$=((2017^{4})^{404}\;mod\;10*2017^{1}\;mod\;10)\;mod\;10$
$=(1*2017^{1}\;mod\;10)\;mod\;10$ ( $\because$ when $a,n$ are co-primes then $a^{\Phi (n)}mod\;n=1$. Here $2017,10$ are co-primes and $\Phi (10)=4$. So, $2017^{4}\;mod\;10=1$ ) $[\Phi(n) = Euler \;\;Totient\;\; Function ]$
$=(2017\;mod\;10)\;mod 10$
$=7\;mod\;10$
$= 7$
option c is the answer..