Proving $n! = n \Rightarrow (n = 1 \quad or \quad n = 2)$ I want to know whether my proof is correct. Any elegant proofs are welcome.


  $n\in\mathbb{N}.\quad$Prove $ (n!=n) \Rightarrow (n=1\quad or\quad n=2)$


$ (n!=n) \Rightarrow (n=1\quad or\quad n=2) \equiv$

$(n\neq1\quad and\quad n\neq2) \Rightarrow (n!\neq n) \equiv$

$n&gt;2 \Rightarrow (n!\neq n)$

Suppose $n&gt;2$. Suppose $n!=n$.

Then $n!=n \equiv $

$(n-1)!=1 \equiv$ 

$(n-1)(n-2)!=1 \quad, (n-2)!$ is defined since $n&gt;2$

Thus, $(n-2)! = 1/(n-1)$

but $(n-2)!\in\mathbb{N}$ and $1/(n-1)\in\mathbb{Q}\setminus\mathbb{N}$, so contradiction.

Therefore $ (n!=n) \Rightarrow (n=1\quad or\quad n=2)$
If $n!&gt;n$, then
$$
(n+1)! = (n+1)n! &gt; (n+1)n &gt; n+1
$$
as long as $n&gt;1$. Now, $3!&gt;3$, and by induction we have proved that $n!&gt;n$ for all $n \geq 3$. Since $1!=1$ and $2!=2$, the assertion is proved.
This seems shorter:

If $n = 1$ or $n = 2$, then  $n! / n = 1$ and thus $n! = n$. 

If $n &gt; 2$, then $n! / n = (n-1)! \geq 2$ and thus $n! \neq n$.

Therefore, $n \in \mathbb{N} \backslash \{ 0 \}$, $n! = n \Rightarrow n = 1 \vee 2$.