I have been using this expression for many years without encountering a counter-example, but I can't take it beyond $k=10000$, so I thought a proof would ease my concerns.
For $k\geq 1\land 2\leq n\leq k+2,$
$$\frac{\binom{k n}{k}^2}{k! (k n)!}\equiv\frac{(k n)!}{(k!)^3 ((k n-k)!)^2}=\frac{p_i\cdot p_{i+1}\cdots}{\text{very large number}},$$
where numerator $p_i\cdot p_{i+1}\cdots$ is the product of primes in $(kn-k,kn].$
The LHS simplifies to the center expression, to which we will refer for the remainder of this discussion.
Using this expression to prove that the numerator will always contain at least one prime is beyond known techniques. But there is one aspect that needs a proof --- is the denominator large enough to cancel all primes $< kn-k$ which are in the numerator?
What would be a good approach to proving this? A pointer to appropriate literature will help, too.
Edit fixed prime indexes.
Edit after reading @GrumpyParsnip's suggestion, I was able to simplify my expression to: $ \frac{(k n)!}{k! (k n-k)! \left\lfloor \frac{k n}{2}\right\rfloor !}$, which I can prove using the Sylvester-Schur proof by Erdos.
My suggestion for proving this is to use Legendre's formula for the power of primes $p$ in a factorial.