What is the last digit of $2017 7^{2017}$?
It is enough to look at the last digit of $7^{2017}$. We notice that the powers of 7 are periodic modulo 10 (i.e., the value of the last digit of $7^{n}, n \geq 1$, is periodic), with a period of 4: the last digit is successively $7,9,3,1,7,9,3,1$, etc. 2016 is a multiple of 4, so the last digit of $7^{2016}$ is a 1. Therefore, that of $7^{2017}$ is a 7.