We write down the first few powers of 7 and notice that
$$7^1=7,\;\;7^2=49,\;\;7^3=243,\;\;7^4=2401,\;\;7^5=16807,\;\;7^6=117649,\;\;7^7=823543,\;\;7^8=5764801,\cdots$$
There is a pattern of the last digit of the powers of 7, as they keep repeating themselves in the sequence 7, 9, 3, and 1.[TABLE="width: 300"]
[TR]
[TD]Last digit of $$7^n$$[/TD]
[TD]7[/TD]
[TD]9[/TD]
[TD]3[/TD]
[TD]1[/TD]
[/TR]
[TR]
[TD]n[/TD]
[TD]1[/TD]
[TD]2[/TD]
[TD]3[/TD]
[TD]4[/TD]
[/TR]
[TR]
[TD][/TD]
[TD]5[/TD]
[TD]6[/TD]
[TD]7[/TD]
[TD]8[/TD]
[/TR]
[TR]
[TD][/TD]
[TD]...[/TD]
[TD]...[/TD]
[TD]...[/TD]
[TD]...[/TD]
[/TR]
[TR]
[TD][/TD]
[TD]...[/TD]
[TD]...[/TD]
[TD]823543[/TD]
[TD]...[/TD]
[/TR]
[TR]
[TD][/TD]
[TD]...[/TD]
[TD]...[/TD]
[TD]...[/TD]
[TD]...[/TD]
[/TR]
[/TABLE]
And our aim is to find the last digit of the number $$7^{7^7}$$, i.e. $$7^{823543}$$.
And this can be found out by dividing the number 823543 by 4 and then judging the location of it in the above table by looking at its remainder.
$$\frac{823543}{4}=205883\frac{3}{4}$$
Hence, we know that $$7^{7^7}=7^{823543}$$ lies in the third column in the table, which has 3 as its last digit.