Easiest to just take the units number 7 from 57 and then test exponents of this. 7^1 = 7, 7^2 = 49, 7^3 = 343, 7^4 = 2401, 7^5 =16807.

Once you get to 7^5, you notice the pattern for the units, 7...9...3...1...7. So regardless of the value of the exponent, the result for the units will be one of those 4 values.

The answers are 7, 9, 3, 1.

(You can test this by raising 57^2 and getting the answer 2349, which has the same tens and units number as 7^2. The same is for 57^3 = 185193)

Hi
power \(x^2\) means x multiplied two times x*x
So \(57^n\) means 57*57*57...n times

Each digit has a cyclicity when multiplied x times
So 7 has cyclicity which repeats after every 4th term
7^1=7..7^2=7*7=9....7^3=7*7*7=9*7=3...7^4=7*7*7*7=3*7=1
So cyclicity is 7,9,3,1,7....


Also it is multiple of odd, so all even are out..
5 as units digit requires 5 as a multiple but here we have 7..