- #1

- 664

- 3

## Homework Statement

Show that

[tex]\forall n \in \mathbb{N}:~~(n!)^{2} < 2^{n^{2}}[/tex]

## The Attempt at a Solution

(1) Show that it is true for n = 1:

(1!)

^{2}= 1; 2

^{1}= 2; => 1 < 2

(2) Show that if it is true for n = p, then it is true for n = p + 1:

Assume that

[tex](p!)^{2} < 2^{p^{2}}[/tex]

Now,

[tex]((p+1)!)^{2} = ((p+1)(p!))^{2} = (p+1)^{2}(p!)^{2}[/tex]

So if it could be shown that

[tex](p+1)^{2}(p!)^{2} < 2^{(p+1)^{2}}[/tex]

then (2) has been demonstrated.

[tex]2^{(p+1)^{2}} = 2^{p^{2}} \cdot 2^{2p} \cdot 2[/tex]

Our assumption shows that

[tex](p!)^{2} < 2^{p^{2}}[/tex]

so I just need to show that the factor [itex](p+1)^{2}[/itex] is less than the factor [itex]2^{2p} \cdot 2[/itex]. I'm not sure how to go about doing that.