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

Question

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)$

Answer 1

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.

Answer 2

This seems shorter:

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

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

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

Answer 3

To prove that $n>2\implies n!>n$ you can use induction.

Base case: $3!=6>3$

Induction step: if $n!>n$ and $n>2$ then $(n+1)!=n!(n+1)>n(n+1)>n+1$.

Answer 4

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 $1$.