How to Prove n! is unbounded from above

Question

I want to show that n! is an unbounded sequence from above. I have that the definition of an unbounded sequence is--

A sequence Xn is unbounded from above if for all m ∈ R, there exists an n ∈ N s.t. Xn > m.

Will setting m = 0 make this statement hold true? I am thinking 0 because n! is never 0 or negative. Let me know if this is the right way to think about this, thank you.

Answer 1

Let $n>2$

We can see $3!=6\gt3$

Now assume $n!>n$

$$ n!>n\Rightarrow n!+n\cdot n!\gt n+1\Rightarrow(n+1)!>n+1 $$

So we have by induction for all $n>2\quad n!>n$

We can see that for any large $n\quad T_n>n$

So the sequence is unbounded above

Note: $n!>n$ is pretty obvious but i figured you wanted a rigorous

Answer 2

Let $m\in\mathbb{R}.$ Define $n=|\lceil m \rceil|+1,$ and you can see that $0<n\in\mathbb{N}.$ We get that $$n!\geq n>|\lceil m \rceil|\geq \lceil m \rceil\geq m.$$