Without direct evaluation- determine which of the following two quantities is larger:
(24!)2 vs 2424
Generalize 24 to n. Then I will analytically prove (n!)^2 > n^n for all natural numbers n>3.
Rearrange the inequality into (n!) * (n!) / (n^n) > 1.
Expand n! two ways, first as an ascending product then as a descending product. Expand n^n as a product of n's. Then the inequality becomes
(1*2*3*...*n) * (n*(n-1)*(n-2)*...*1) / (n*n*n*...*n) > 1
Now group the kth terms of each expansion to create a new form
(1*n)/n * (2*(n-1))/n * ... * (k*(n-k))/n * ... (n*1)/n > 1
The first and last terms of this expansion are each equal to one, so all that is left is to show all the interior terms are greater than one.
Consider the arbitrary term (k*(n-k))/n and apply some algebra.
(k*(n-k))/n > 1
nk - k^2 > n
n*(k-1) > k^2
n > k^2/(k-1) = k+1 - 1/(k-1)
n-1 > k - 1/(k+1)
k ranges from 2 to n-1 thus we know n-1>=k so then the derived inequality is also true. Therefore every intermediate term is in fact greater than 1. Which then implies the original inequality (n!) * (n!) / (n^n) > 1 is true.
Solution
