Question 170682
Let's write the equation
{{{(7^7)^7}}}

Using the power of powers rule, that yields
{{{7^49}}}

Now, 7*7 = 49
49*7 = 343
343*7 = 2401
2401 * 7 = 16807

Look at the ones digit above. You can see the pattern 7,9,3,1,7...
So the pattern repeats after each group of 4.
Thus 7^(4x+1) ends in 7
7^(4x+2) ends in 9
7^(4x+3) ends in 3
7^(4x+4) ends in 1

49/4 = 12.25 = (4*12) + 1
That means there are 12 groups of 4, plus one left over. That means you will be left with the ones digit ending in 7