$7^0 = 1 , 7^1 = 7 , 7^2 = 49 , 7^3 = 343 , 7^4 = 2401 , 7^5 = 16807$
now if we see only the last digit than we can say for
$7^0 = 1 , 7^1 = 7 , 7^2 = 9 , 7^3 = 3 , 7^4 = 1 , 7^5 = 7$
so we can say that after every forth multiple of 7 , its last digit keeps on repeating in cycle .so if , we get to find last digit of 7 raise to x than we should divide x by 4 and see what is the remainder of it (or) we should do x%4 and we should write the last digit of the respective value.
as x=19522 hence x%4 = 2 therefore
last digit of 7 raise to 19522 = $7^2 = 9$ and hence the answer will be 9.