We seem to have seen this one twice before: here and here.

Maxal wrote: \[ \prod_{k=1}^n k^{2k-n-1} = \dfrac{n!^{n-1}}{1!^2 2!^2\dots (n-1)!^2} = \prod_{i=1}^{n-1} {n\choose i} \] is a product of integers.

can you give a brief proof without induction?

Notice that $$\prod_{k=1}^{n}{k^{2k-n-1}}=\frac{\prod_{k=1}^{n}{k^{2k}}}{\prod_{k=1}^{n}{k^{n+1}}}=\frac{\left(\prod_{k=0}^{n-1}{\frac{n!}{k!}}\right)^2}{n!^{n+1}}=\frac{n!^{n-1}}{\prod_{k=1}^{n-1}{k!(n-k)!}}=\prod_{k=1}^{n-1}\binom{n}{k},$$so we are done. $\square$

The product is equivalent to $$\frac{(1^1\cdot 2^2\cdots n^n)^2}{(n!)^{n+1}}=\frac{(\frac{n!}{0!})^2(\frac{n!}{1!})^2\cdots(\frac{n!}{(n-1)!})^2}{(n!)^{n+1}}=\frac{(n!)^{n-1}}{(1!2!\cdots(n-1)!)^2}.$$ We see that this is an integer because it is equivalent to $$\binom{n}{1}\binom{n}{2}\cdots\binom{n}{n-1}$$ as desired.

Write down Formulas! We simply grind through the algebra. \begin{align*} \prod^n_{k=1} k^{2k-n-1} & = \frac{1}{(n!)^{n+1}} \cdot \prod_{k=1}^n k^{2k} = \frac{1^2 \cdot 2^4 \cdot \cdots \cdot n^{2n}}{(n!)^{n+1}}\\ & = \frac{(n!)^2\left(\frac{n!}{1!}\right)^2\left(\frac{n!}{2!}\right)^2 \cdots \left(\frac{n!}{{n-1}!}\right)^2}{(n!)^{n+1}}\\ &= 1\cdot \left(\frac{n!}{(1!)^2}\right) \left(\frac{n!}{(2!)^2}\right) \cdots \left(\frac{n!}{(n-1)!^2}\right)\\ &= \frac{n!^{n-1}}{(1!2!\dots (n-1)!)^2}\\ &= \frac{n!}{1!(n-1)!} \cdot \frac{n!}{2!(n-2)!} \cdot \cdots \cdot \frac{n!}{(n-1)!1!}\\ &= \prod_{k=1}^{n-1}\binom{n}{k} \end{align*}which is a product of integers and hence is an integer itself.