Prove that: $2^n < n!$ Using Induction

Question

I'm told to show that $2^n < n!$ using induction

This is my attempt at it:

BC: $n=4, 2^4 = 16 < 4!$

IH: n = k, $2^k < k!$

IS: try n = k+1

I'm told to only work from one side, so I try the left side:

2^(k+1)

But I'm stuck here, any ideas?

Answer 1

It is true only for $n\ge 4$.

Hint for the inductive step:

$2^{k+1}=2\cdot2^k<2\cdot k!$, so it is enough to prove $2\cdot k!\le (k+1)!$.

Answer 2

You have $2^k<k!$ by induction hypotheses. Then $2^k(k+1)<(k+1)k!=(k+1)!$, if $k>3$ you have that $2^{k+1}<(k+1)2^k$ and the result follows.

Answer 3

This is true for $n \geq 4$.

Base step: $2^4 < 4!$ i.e. $16 < 24$ true!

Induction Step: suppose $2^{n-1} < (n-1)!$. Since $2<n$ you can multilpy the left side for $2$ and the right side for $n$ keeping the inequality true, so you get

$2 \cdot 2^{n-1} < n(n-1)!$ i.e. $2^n < n!$.