Raising self to higher powers

Working modulo 5 in the base and modulo $4=\phi(5)$ in the exponent, we see that the given number is congruent to $2^1=2$ modulo 5. Since the number is odd, the last digit must be a $\boxed{7}$ .

Using cyclicity of 7 we can easily get the answer as 7

We see a pattern of 7 as when raised to the power 1, its last digit is 7. When 2, it is 9 For 3, it is 3 and for 4, it is 1. And then it forms a pattern. When the given no. Is divided by 4 it gives remainder 1[ (ak+1)/a) gives 1 remainder]. Now since remainder is one, we know that the last digit is 7. Please correct me if I am wrong.