(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Show that

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

3. 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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Induction Again

**Physics Forums | Science Articles, Homework Help, Discussion**