Question

Prove that (n!)! is divisible by (n!)^(n-1)!

Answer 1

Answer:

Step-by-step explanation:

P=(n!)!(n!)(n−1)!P=(n!)!(n!)(n−1)! as (n!)!(n!)(n!)⋯(n!)(n!)!(n!)(n!)⋯(n!)

Number of n!n!s in the denominator is (n−1)!(n−1)!

Hope it will help you

Answer 2

$\huge\blue{Answer}$

P=(n!)!(n!)(n−1)!P=(n!)!(n!)(n−1)! as (n!)!(n!)(n!)⋯(n!)(n!)!(n!)(n!)⋯(n!)

Number of n!n!s in the denominator is (n−1)!(n−1)!

HOPE IT HELPS U !!