n to the n is greater than n factorial squared (idea)

Proof:

nⁿ = 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 - i² + 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 nⁿ.

QED