If 7 divides `32^(32^(32))`, then find the remainder

Explanation

We have , ` 32 = 2^(5)`
` therefore 32^(32) = (2^(5))^(32) = 2^(160) = (3-1)^(160)`
` = ""^(160)C_(0) (3)^(160) - ""^(160)C_(1) (3) ^(159) + …-""^(160)C_(159) (3) + 1 `
` 3(3^(159) - ""^(160)C_(159)) (3)^(158) + ...- ""^(160)C_(159)) + 1`
` = 3m + 1, m in I^(+)`
Now , ` 32^(32^(32)) = 32^(3m +1) = 2^(5(3m +1)) = 2^(15 m + 5)`
` = 2^(2) * 2^(3(5m + 1)) = 4 (8)^(5m + 1) = 4(7 +1)^(5m +1)`
`= 4[ ""^(5m +1)C_(0) (7)^(5m+1) + ""^(5m+1)C_(1) + (7)^(5m)+""^(5m+1)C_(2) (7) ^(5m-1) + ...+ ""^(5m +1)C_(5m)(7 )+1]`
`= 4[7( ""^(5m +1)C_(0) (7)^(5m) + ""^(5m+1)C_(1) (7)^(5m-1)+..+""^(5m-1)C_(5m))+1]`
` = 4 [ 7k + 1 ] ` , where k is positive integer = 28k + 4
` therefore (32^(32^(32)))/(7) = 4k + (4)/(7)`
Hence , the remainder is 4 .