TITLE: Proving $n! = n \Rightarrow (n = 1 \quad or \quad n = 2)$
QUESTION [5 upvotes]: 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$
$(n\neq1\quad and\quad n\neq2) \Rightarrow (n!\neq n) \equiv$
$n>2 \Rightarrow (n!\neq n)$
Suppose $n>2$. Suppose $n!=n$.
Then $n!=n \equiv $
$(n-1)!=1 \equiv$ 
$(n-1)(n-2)!=1 \quad, (n-2)!$ is defined since $n>2$
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)$

REPLY [3 votes]: If $n!>n$, then
$$
(n+1)! = (n+1)n! > (n+1)n > n+1
$$
as long as $n>1$. Now, $3!>3$, and by induction we have proved that $n!>n$ for all $n \geq 3$. Since $1!=1$ and $2!=2$, the assertion is proved.

REPLY [3 votes]: Obviously, $n\ne0$, and we can simplify:
$$(n-1)!=1.$$
By inspection, $n-1=0$ and $n-1=1$ are solutions. All other factorials have a factor that makes them exceed $1$.