Explanation

(c)
We have, $32^{32} = {(2^{5})}^{32} = 2^{160} = {(3 - 1)}^{160}$
$= \ ^{160}C_{0}3^{160} - \ ^{160}C_{1} \bullet 3^{159} + ... - \ ^{160}C_{159} \bullet 3 + \ ^{160}C_{160}3{^\circ}$
$= 3m + 1,\ where\ m \in N$
$32^{(32)^{(32)}} = (32)^{3m + 1}$
$= {(2^{5})}^{3m + 1} = 2^{15m + 5}$
$= 2^{3(5m + 1)} \bullet 2^{2} = {(2^{3})}^{5m + 1} \bullet 2^{2}$
$= (7 + 1)^{5m = 1} \times 4$
$= \{\ ^{5m + 1}C_{0}7^{5m + 1} + \ ^{5m + 1}C_{1}7^{5m} + ... + \ ^{5m + 1}C_{5m + 1}7 + \ ^{5m + 1}C_{5m + 1}{.7}^{0}\} \times 4$
$= (7n + 1) \times 4$,
where $n = \ ^{5m + 1}C_{0}7^{5m + 1} + ... + \ ^{5m + 1}C_{5m} \bullet 7$
$28n + 4$
Thus, when $32^{{(32)}^{(32)}}$ is divided by 7, the remainder is 4