How do you work out this Euler's Theorem problem?

The problem is find the last digit of 7^77^777





So 7 which is to the 77 and the 77 is to the 777 (mod 10) basically





I know it is an Euler's Theorem problem but can somebody show how to work it through to find the last digit.How do you work out this Euler's Theorem problem?
I think cdmillstx may have an error in his math. My approach to the problem is:





As cdmillstx says, the last two digits as you increase the powers of 7 are: 07, 49, 43, 01, and then you repeat the cycle.





However, 777 = 1 mod 4, not 3 mod 4. So, the last two digits of 7^777 are the same as 7^1, i.e. 07.





As cdmills says, the last two digits as you increase the powers of 11 are 11, 21, 31, 41, etc., repeating every 10. So, the last two digits of 11^777 are 71.





So, to get the last two digits of 77^777, get the last two digits of 7^777 * 11^777 = mod 100(07 * 71) = 97.





Finally, since 97 = 1 mod 4, the last two digits of 7^97 are the same as 7^1, i.e. 07.How do you work out this Euler's Theorem problem?
I'm not versed in Euler's Theorem, but:





From the sequence of powers of 7:


1, 7, 49, 343, 2401, ....


We see that the last two digits cycle through 01, 07, 49, 43


Thus, 7 ^ 777 will have the same last two digits as 7 ^ (777 mod 4), or 77 ^ 3, which is 43.


Pascal's triangle shows that the last two digits of 11 ^ 777 is 71.


7^777 * 11^777 = 77^777, so the last two digits of 77^777 are the last two digits of 43*71, or 53.


Employ the 7^n logic of the last two digits again and we see the the last two digits of 7^(77^777) is the same as the last two digits of 7^(53 mod 4), or 3.





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Oops, Zanti's right about my goof.