Take last digit of 97 i.e. 7
Now $7\times 7=9, 9\times 7=3,3\times 7=1,1\times 7=7$ (taking last digit only)
So, 7 is a cycle of 4 and after multiplication of four 7 we get 7 again. i.e., $7^{4n+1}$ will have unit digit as 7 for $n \geq 0$.
$2013=503\times 4+1$
Now, last digit of $97^{2013}$
= Last digit of $7^{503\times 4+1} =7.$