I am aware that
$$\frac{1}{\Gamma\left(1-x\right)\Gamma\left(1+x\right)}\prod_{n=1}^{m}\frac{1}{1-\left(\frac{x}{n}\right)^{2}}=\frac{\Gamma(m+1)^{2}}{\Gamma\left(m+x+1\right)\Gamma\left(m-x+1\right)}\tag{1}$$
From this post I derived a similar-looking identity:
$$\prod_{n=1}^{m}\frac{\Gamma\left(n\right)^{2}}{\Gamma\left(n+x\right)\Gamma\left(n-x\right)}\cdot\frac{1}{\left(1-\left(\frac{x}{n}\right)^{2}\right)^{n}}=\left(\frac{\Gamma\left(m+1\right)^{2}}{\Gamma\left(m+x+1\right)\Gamma\left(m-x+1\right)}\right)^{m}\tag{2}$$
My proof works, but I feel like there should be a more obvious proof. I've tried using $(1)$ to get to $(2)$, but keep getting it wrong. I would take a hint.
Let's denote (for brevity) $$F_n(x)=\frac{\Gamma(n)^2}{\Gamma(n+x)\Gamma(n-x)},\qquad f_n(x)=\frac{1}{1-(x/n)^2},$$ so that $(1)$ reads $F_{m+1}(x)=F_1(x)\displaystyle\prod_{n=1}^{m}f_n(x)$. Now the LHS of $(2)$ is equal to \begin{align*} \prod_{n=1}^m\big(F_n(x)f_n(x)^n\big) &=\prod_{n=1}^m\left(f_n(x)^n F_1(x)\prod_{k=1}^{n-1}f_k(x)\right) \\&=F_1(x)^m\times\prod_{n=1}^{m}f_n(x)^n\times\prod_{n=1}^{m}\prod_{k=1}^{n-1}f_k(x) \\\color{gray}{[\ 1\leqslant k<n\leqslant m\ ]}\quad&=F_1(x)^m\times\prod_{n=1}^{m}f_n(x)^n\times\prod_{k=1}^{m-1}\prod_{n=k+1}^{m}f_k(x) \\&=F_1(x)^m\times\prod_{n=1}^{m}f_n(x)^n\times\prod_{k=1}^{m-1}f_k(x)^{m-k} \\&=F_1(x)^m\prod_{n=1}^{m}f_n(x)^m=F_{m+1}(x)^m, \end{align*} equal to the RHS of $(2)$ as expected.