Prove that: $2^n < n!$ Using Induction [duplicate]

Solution 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 $2\cdot k!\le (k+1)!$.

Solution 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.

Solution 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. $2^n < n!$.