Here we show the claim is valid for odd positive integers $n$. For even $n$ it can be shown analogously.
We start with the denominator, the somewhat easier part.
We obtain for odd integers $n>1$
\begin{align*}
\color{blue}{\prod_{k=1}^{n-1}(2k+1)^{n-k}}&=3^{n-1}\cdot5^{n-2}\cdots(2n-3)^2(2n-1)\\
&=\prod_{k=1}^{n-1}\frac{(2n+1-2k)!}{(2n-2k)!!}\tag{1}\\
&=\prod_{k=1}^{n-1}\frac{(2k+1)!}{(2k)!!}\tag{2}\\
&\,\,\color{blue}{=\prod_{k=1}^{n-1}\frac{(2k+1)!}{2^kk!}}\tag{3}
\end{align*}
Comment:
In (1) we rearrange the right-hand side of the first line by using factorials and double-factorials $(2p)!!=(2p)(2p-2)\cdots4\cdot2$.
In (2) we change the order of the factors of the numerator by setting $k\to n-k$.
In (3) we use the identities $(2p)!=(2p)!!(2p-1)!!$ and $(2p)!!=2^pp!$.
Now we take a look at the numerator
\begin{align*}
\color{blue}{n^{\lceil \frac{n}{2}\rceil}}&\color{blue}{\prod_{k=1}^{n-1}\frac {(n^2-k^2)^{\lfloor \frac{n-k+1}{2}\rfloor}}{(2k+1)^{n-k}}}\\
&=\left(\prod_{k=1}^{n-1}\frac {k^{\lfloor \frac{n-k+1}{2}\rfloor}}{(2k+1)^{n-k}}\right)n^{\lceil \frac{n}{2}\rceil}
\left(\prod_{k=1}^{n-1}\frac {(n-k)^{\lfloor\frac{n-k+1}{2}\rfloor}}{(2k+1)^{n-k}}\right)\tag{4}\\
&=1^1\cdot2^1\cdot3^2\cdot4^2\cdots(n-2)^{\frac{n-1}{2}}(n-1)^{\frac{n-1}{2}}n^{\frac{n+1}{2}}\\
&\qquad\cdot (n+1)^{\frac{n-1}{2}}(n+2)^{\frac{n-1}{2}}\cdots(2n-4)^2(2n-3)^2(2n-2)^1(2n-1)^1\tag{5}\\
&=(2n-1)!\cdot\frac{(2n-3)!}{2!}\cdot\frac{(2n-5)!}{4!}\cdots\frac{n!}{(n-2)!}\tag{6}\\
&\,\,\color{blue}{=n!\prod_{k=1}^{\frac{n-1}{2}}\frac{(2n+1-2k)!}{(2k)!}}\tag{7}
\end{align*}
Comment:
In (4) we use $n^2-k^2=(n-k)(n+k)$ and put the factor $n^{\lceil \frac{n}{2}\rceil}$ in the middle. We also change the order of the factors of the numerator at the left-hand product by $k\to n-k$.
In (5) we use a more descriptive style for the expression (4).
In (6) we collect terms to factorials starting with the right-most factor in steps of two terms. We compensate the missing factors at the left-hand side by multiplying with factors $\frac{1}{(2k)!}$ accordingly.
In (7) we write the expression (6) more compactly using the product symbol.
We can now take the results (3) and (7) and obtain
\begin{align*}
\color{blue}{n^{\lceil \frac{n}{2}\rceil}}&\color{blue}{\prod_{k=1}^{n-1}\frac {(n^2-k^2)^{\lfloor \frac{n-k+1}{2}\rfloor}}{(2k+1)^{n-k}}}\\
&=n!\prod_{k=1}^{\frac{n-1}{2}}\frac{(2n+1-2k)!}{(2k)!}\prod_{k=1}^{n-1}\frac{2^kk!}{(2k+1)!}\tag{8}\\
&=2^{\binom{n}{2}}\prod_{k=1}^{\frac{n-1}{2}}\frac{1}{(2k)!}\prod_{k=1}^{\frac{n-3}{2}}\frac{1}{(2k+1)!}\prod_{k=1}^{n-1}k!\tag{9}\\
&\,\,\color{blue}{=2^{\binom{n}{2}}}\tag{10}
\end{align*}
and the claim follows.
Comment:
In (8) we take the result (3) and multiply it with the reciprocal of (7).
In (9) we cancel $\prod_{k=1}^{\frac{n-1}{2}}(2n+1-2k)!$ with $\prod_{k=\frac{n-1}{2}}^{n-1}\frac{1}{(2k+1)!}$ and factor out $\prod_{k=1}^{n-1}{2^k}=2^{\sum_{k=1}^{n-1}k}=2^{\binom{n}{2}}$.
In (10) we cancel everything besides the power of $2$.
Note: In order to show the case when $n$ is even we can adapt this answer starting from (5).
