<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] A more thorough version of the answer given by fleablood, avoiding assumptions and unexamined cases.
The stated question, find all solutions for $n$ such that $n^2=(n-1)!+1$
Since for all non-negative integers, $k!\ge 1$, we conclude $n^2\ge 2 \Rightarrow n>1 \Rightarrow (n-1)>0$
Rearranging, $(n-1)!=n^2-1=(n-1)(n+1)$
Restating, $(n-1)(n-2)!=(n-1)(n+1)$
Since $(n-1)>0$, we are allowed to divide through by $(n-1)$ and obtain $(n-2)!=(n+1)$. And since $n>1 \Rightarrow n+1 > 2$, the largest integer in the factorial must be at least as large as $2$, viz: $n-2 \ge 2$
Therefore $n-2$ must have at least one divisor $d>1$. But $d\mid (n-2) \Rightarrow d\mid (n-2)! \Rightarrow d\mid (n+1)$
Any factor which divides each of $(n-2)$ and $(n+1)$ must divide their difference, $(n+1)-(n-2)=3$
The only factor greater than $1$ which divides $3$ is $3$ itself, and since $(n-2)$ has no other common factors with $(n+1)$, it must be the case that $n-2=3 \Rightarrow n=5$ is the sole solution to the stated problem.
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