Factorials

$\displaystyle \begin{aligned} \frac{(n^2 + 1)!}{(n^2 - 1)!} & = 10n^2 \\ \frac {(n^2 + 1) \cdot (n^2) \cdot (n^2 - 1)!}{(n^2 - 1)!} & = 10n^2 \\ (n^2 + 1)(n^2) & = 10n^2 \\ n^2 + 1 & = 10 \\ n & = \pm 3 \end{aligned}$

$n > 0 \ \ \Longrightarrow \ \ \boxed{n = 3}$

$\large \frac { ({ n }^{ 2 }+1)! }{ ({ n }^{ 2 }-1)! } =10{ n }^{ 2 }\\ \frac { ({ n }^{ 2 }-1)!({ n }^{ 2 })({ n }^{ 2 }+1) }{ ({ n }^{ 2 }-1)! } =10{ n }^{ 2 }\\ { n }^{ 2 }({ n }^{ 2 }+1)=10{ n }^{ 2 }\\ { n }^{ 2 }+1=10\\ { n }^{ 2 }=9\\ n=3$

let's make $n^2 = m$

So we end up with the follow eq

$\frac{(m + 1)!}{(m - 1)!} = 10m$ let's develop this:

$(m+1)! = m!(m+1)$

$(m-1)! = \frac{m!}{m}$

$\frac{m!(m+1)}{ \frac{m!}{m}} = 10m$ developing this ugly equation we have

$m^2 + m = 10m$

$m^2 - 9m = 0$

now divide everything by m or solve the quadratic taking only the positive integer value

$m = 9$

$\sqrt{m}=n=3$