Not true. n

^{n} ≤ n!

^{2} for all n a

natural number.

*Proof:*

n^{n} = n * n * n * ... * n * n * n

n! = n * (n-1) * (n-2) * ... * 3 * 2 * 1

n! = 1 * 2 * 3 * ... * (n-2) * (n-1) * n

=> n!^{2} = (n*1) * ((n-1)*2) * ((n-2)*3) * ... * (3*(n-2)) * (2*(n-1)) * (1*n)

This sequence contains n terms and each term is of the form ((n-i)*(i+1))

(n-i)*(i+1) = ni - i^{2} + n - i

This is obviously greater than or equal to n for all i such that 1 ≤ i ≤ n

Hence, the sequence is the product of n terms, all greater than n. Hence it is greater than or equal to n^{n}.

QED

*Jurph*: Fixed. Now it says ≤ all the way through, rather than <. Stupid typos.