(n+1)! = (n+1)* (n)*(n-1)*........1 = (n+1)*(n!) > n!
Hence n! = O((n+1)!) => option (1) is wrong.
log4n = (1/2)* log2n, So we can write log2n as follows:
(1/2)* log2n ≤ (1/2)* log2n ≤ log2n => (1/2)* log2n = ⊝(log2n) => log4n = ⊝(log2n) Hence (!!) is correct.
√logn ≥ loglogn => loglogn = O(√logn) => (!!!) is wrong.
Hence only (!!) is correct. Hence option (C) is correct.