Odd tower

Cyclicity of 3 is four so divide the given power by 4 5^7^9 By binomial expansion we get ((4+1)^7^9)/4 The remainder will be 1^7^9= 1 So last digit of 3^5^7^9 is same as that of 3^1 i.e. 3 Hence answer is 3

We start from top to bottom: $\text{Last digit of}\space 7^9 \text{is 1(cyclicity of 7 is 7,9,1,1...)}$ $\text{Last digit of}\space 5^1 \text{is 5(cyclicity of 5 is 5,5,5,5...)}$ $\text{Last digit of}\space 3^5 \text{is} \boxed{3}\text{(cyclicity of 3 is 3,9,7,1,3,9,7,1,...)}$