A strange equation?

$n!=k n^3 \\ => (n-1)!=k n^2$ .

It is obvious that no prime will satisfy the above equation because $(n-1)!$ does not contain the prime $n$ .

So, at least $n$ will be a composite number.Let $n=p_1^{k_1} p_2^{k_2}.....p_r^{k_r} \implies n^2=p_1^{2 k_1} p_2^{2 k_2}.....p_r^{2 k_r}$ .So, in the $(n-1)!$ all these factors should be present.Let us assume that $max \{p_1,p_2,p_3,,,,,p_r\}=p_r$ ..To satisfy the equation we must have

$\large 2 k_{r} p_r \leq (n-1)$

With the help of these condition we can easily find solutions in the given range.