Is there any number which $n!$ is lower than $2^n$ or same?

Question

I interested in this question.
how many numbers meet this condition?
I think a few of them meet this but I want a proof for this.
also I'm not very pro in mathematics.

Answer 1

Observe that $n!<2^n$ if $n<4$ and $n!>2^n$ for $n=4$

Let $m!>2^m$ for $m\ge 4$

Then $(m+1)!>(m+1)2^m>2^{m+1}$ if $m+1\ge 2\iff m\ge1$ which is true as $m\ge 4$

Answer 2

This is a good time to just "plug-and-play" with values of $n$. There will be only a handful that meet this condition:

Namely, when and only when $n = 0, 1, 2, 3$, then we have $n! \leq 2^n$. For a valid proof of this, plug in and test out $n = 0, n = 1, n = 2, and n = 3$ to confirm.

Challenge: Use proof by induction on $n$ to prove that for all values of $$n \geq 4 \implies n! \gt 2^n$$

Answer 3

This condition will hold for all $n$ from $0$ up to $3$, then $n! > 2^n$ for all $n > 3$

Answer 4

You can see the Gamma function $\Gamma(x+1)$ (which is the extension to the reals of the factorial $x!$) intercepts $2^x$ at $x\approx 3.46$. After this it is always greater.