If $$x_{n}=x_{n-1}+\sqrt{x_{n-1}}+1, x_{0}=1$$ so we can say, that $$\prod\limits_{n=1}^{\infty}\left(1-\frac{1}{x_{n}^{3/2}}\right)^{-1}=\left[\left(1-\frac{1}{3^{3/2}}\right)\left(1-\frac{1}{(4+\sqrt{3})^{3/2}}\right)\left(1-\frac{1}{(5+\sqrt{3}+\sqrt{4+\sqrt{3}})^{3/2}}\right)\cdots\right]^{-1}=\frac{3}{2}$$ All square roots are positive.
How can I prove it?
Let $u_n = \sqrt{x_n}$, we have
$$u_n^2 = u_{n-1}^2 + u_{n-1} + 1 = \frac{u_{n-1}^3 - 1}{u_{n-1} - 1} \implies u_{n-1}^3-1 = u_n^2(u_{n-1}-1) $$ This leads to
$$1 - \frac{1}{x_n^{3/2}} = \frac{u_n^3-1}{u_n^3} = \frac{u_{n+1}^2(u_n-1)}{u_n^3} = \frac{u_{n+1}^2(u_n^2-1)}{u_n^3(u_n+1)} = \frac{u_{n+1}^2(u_n^2-1)}{u_n^2(u_{n+1}^2-1)}\\ = \left.\frac{x_n - 1}{x_n}\right/\frac{x_{n+1}-1}{x_{n+1}} $$ The product at hand is a telescoping one with partial products
$$\prod_{n=1}^p \left(1 - \frac{1}{x_n^{3/2}}\right)^{-1} =\left.\frac{x_1}{x_1 - 1}\right/\frac{x_{p+1}}{x_{p+1}-1}$$
Since $x_n = x_{n-1}+\sqrt{x_{n-1}} + 1 \ge x_{n-1} + 1$ and $x_0 = 1$, we can show $x_n \ge n + 1$ by induction. This means $$\lim\limits_{p\to\infty} x_p = \infty \implies \lim\limits_{p\to\infty}\frac{x_{p+1}}{x_{p+1}-1} = 1$$ As a result, $$\prod_{n=1}^\infty \left(1 - \frac{1}{x_n^{3/2}}\right)^{-1} = \frac{x_1}{x_1 - 1} = \frac{3}{3-1} = \frac32$$