Find sum of digits of:(1999)^1999.
[The final answer should be a single digit number, for example, (2)^16 = 65536 and the sum of its digits will be given by (6 + 5 + 5 + 3 + 6 = 25, which again will be reduced to 2 + 5 = 7].

Let f(x): sum of digits of x to a single digit. Let g(x)=x mod 9 then f(x)=g(x); if g(x)>0 =9; if g(x)=0 we have to find g(1999^1999)=1999^1999 mod 9 = ((1999)mod 9)^1999 mod 9 = 1. As g(x)>0, f(x)=1 So, sum of digits of (1999)^1999=1.