70 = 1 (consider that as 01)

71 = 7 (consider that as 07)

72 = 49

73 = 343

74 = 2401

75 = 16807

76 = 117649

77 = 823543

78 = 5764801

...


So we see that the last two digits of powers of 7 are

01, 07, 49, 43, 01, 07, 49, 43, 

ad the last two digits continue to repeat in groups of 4 forever. 

So to find the last two digits of any positive integral power of

7, we can just divide the power by 4 and take only the remainder.

If the remainder is 0, the last two digits are 01,
If the remainder is 1, the last two digits are 07,
If the remainder is 2, the last two digits are 49,
If the remainder is 3, the last two digits are 43,

To find 71997, we divide 1997 by 4

          499  
       4)1997
         16
          39 
          36
           37 
           36
            1

The remainder is 1, so the last two digits of 71997 are 07.

Edwin