<s>[INST] Find all $n≥1$ natural numbers such that : $n^{2}=1+(n-1)!$ Problem :
Find all $n≥1$ natural numbers such that : $n^{2}=1+(n-1)!$
My try : 
$n=1$ we find : $1=1+1$ $×$ 
$n=2$ we find : $4=1+1$ $×$ 
$n=3$ we find : $9=1+2$ $×$ 
$n=4$  we find : $16=1+6$ $×$
$n=5$ we  find : $25=1+24$ $√$ 
Now how I prove $n=5$ only the solution ?
 [/INST] Hint If $n \geq 6$ then 
$$1+(n-1)!  \geq 1+2 \cdot 3 \cdot (n-2) \cdot (n-1)$$
Show that 
$$2(n-1) \geq n \\
3(n-2) \geq n$$
for $n \geq 6$.
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