I am wondering if the following finite product has a nice closed form. This appears to be related to this question (if the numerator and denominator can be put into closed form seperately), but this one seems more complicated. Unfortunately this is not really my area, so I don't have much in the way of work to show. Let $n$ and $k$ be fixed positive integers with $k<n$.
$$\prod_{j=1}^k\frac{n+j}{n-j}$$
The product can be written as the ratio of two binomial coefficients: $$\prod_{j=1}^k\dfrac{n+j}{n-j}=\frac{\frac{\prod_{j=1}^k(n+j)}{k!}}{\frac{\prod_{j=1}^k(n-j)}{k!}}=\frac{\binom{n+k}{k}}{\binom{n-1}{k}}.$$ Does it help?
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
I'll assume $\ds{n \geq k \geq 1}$:
\begin{align} \prod_{j = 1}^{k}{n + j \over n - j} & = \pars{-1}^{k}\, {\prod_{j = 1}^{k}\pars{j + n} \over \prod_{j = 1}^{k}\pars{j - n}} = \pars{-1}^{k}\, {\pars{1 + n}^{\overline{\large k}} \over \pars{1 - n}^{\overline{\large k}}} = \pars{-1}^{k}\, {\Gamma\pars{1 + n + k}/\Gamma\pars{1 + n} \over \Gamma\pars{1 - n + k}/\Gamma\pars{1 - n}} \\[5mm] & = \pars{-1}^{k}\, {\Gamma\pars{1 + n + k} \over \Gamma\pars{1 + n}} {\Gamma\pars{1 - n} \over \Gamma\pars{1 - n + k}} \end{align}
For $\ds{z \not= n}$:
\begin{align} {\Gamma\pars{1 - z} \over \Gamma\pars{1 - z + k}} & = {\pi \over \Gamma\pars{z}\sin\pars{\pi z}}\,{\Gamma\pars{z - k}\sin\pars{\pi\bracks{z - k}} \over \pi} \\[5mm] & = {\Gamma\pars{z - k} \over \Gamma\pars{z}} \,\bracks{\cos\pars{\pi k} - \cot\pars{\pi z}\sin\pars{\pi k}} = {\Gamma\pars{z - k} \over \Gamma\pars{z}}\,\pars{-1}^{k} \end{align}
Lets $\ds{z \to n}$:
\begin{align} \prod_{j = 1}^{k}{n + j \over n - j} & = \bbx{{\Gamma\pars{1 + n + k} \over \Gamma\pars{1 + n}}\, {\Gamma\pars{n - k} \over \Gamma\pars{n}}} = {\pars{n + k}! \over n!}\, {\pars{n - k - 1}! \over \pars{n - 1}!} = {\pars{n + k}! \over k!\,n!}\, {k!\pars{n - k - 1}! \over \pars{n - 1}!} \\[5mm] & = \bbx{{{n + k \choose k} \over {n - 1 \choose k}}} \end{align}
I assume that $n > k$; otherwise there is a divide by zero.
I am also assuming that $n$ and $k$ are integers.
$\begin{array}\\ p(n, k) &=\prod_{j=1}^k\dfrac{n+j}{n-j}\\ &=\dfrac{\prod_{j=1}^k(n+j)}{\prod_{j=1}^k(n-j)}\\ &=\dfrac{\prod_{j=n+1}^{n+k}j}{\prod_{j=n-k}^{n-1}j}\\ &=\dfrac{(n+k)!/n!}{(n-1)!/(n-k-1)!}\\ &=\dfrac{(n+k)!(n-k-1)!}{n!(n-1)!}\\ \end{array} $
For the fun of it, removing the assumption on $n$ being an integer,
$\begin{array}\\ \ln(p(n, k)) &=\sum_{j=1}^k\ln(\dfrac{n+j}{n-j})\\ &=\sum_{j=1}^k\ln(\dfrac{1+j/n}{1-j/n})\\ &=\sum_{j=1}^k(\ln(1+j/n)-\ln(1-j/n))\\ &=\sum_{j=1}^k(\sum_{m=1}^{\infty}\frac{(-1)^{m-1}j^m}{n^m}+\sum_{m=1}^{\infty}\frac{j^m}{n^m})\\ &=\sum_{j=1}^k(\sum_{m=1}^{\infty}\frac{j^m}{n^m}((-1)^{m-1}+1)\\ &=2\sum_{j=1}^k\sum_{m=1}^{\infty}\frac{j^{2m-1}}{n^{2m-1}}\\ &=2\sum_{m=1}^{\infty}\frac1{n^{2m-1}}\sum_{j=1}^kj^{2m-1}\\ &\approx 2\sum_{m=1}^{\infty}\frac1{n^{2m-1}}\frac{k^{2m}}{2m} \qquad\text{(an additional term would make this more accurate)}\\ &=2n\sum_{m=1}^{\infty}\frac1{n^{2m}}\frac{k^{2m}}{2m}\\ &=2n\sum_{m=1}^{\infty}\frac{(k/n)^{2m}}{2m}\\ &=n\sum_{m=1}^{\infty}\frac{((k/n)^2)^{m}}{m}\\ &=-n\ln(1-(k/n)^2)\\ \text{so that}\\ p(n, k) &\approx(1-(k/n)^2)^{-n}\\ \end{array} $
