By the Fundamental Theorem of Algebra, it is easily seen that for a monic polynomial $p(x) \in \mathbb{C}[x]$, $$\prod _{j=1}^n p(j) = \frac{\prod_{p(r)=0}\Gamma(1+n-r)}{\prod_{p(r)=0}\Gamma(1-r)},$$ where the products in the right-hand side of the above equality are over all roots $r$ of $p(x)$. For example, we have that $$\prod_{j=1}^{n} (j^2+1) = \frac{\Gamma (1+n-i) \Gamma (1+n+i)}{\Gamma (1-i) \Gamma (1+i)},$$ letting $i$ denote the imaginary unit.
Mathematica is not able to symbolically evaluate the product $$\prod _{j=1}^n \left(\sqrt{j}+1\right),$$ and it is not obvious to me as to how to evaluate the above product in terms of special functions such as the gamma function. It is natural to ask:
(1) Is there a known way of evaluating the product $\prod _{j=1}^n \left(\sqrt{j}+1\right)$ in terms of special functions such as the gamma function?
(2) More generally, is there a known way of evaluating products of the form $\prod _{j=1}^n a(j)$ in terms of special functions such as the gamma function, letting $a(j)$ denote an algebraic function?
Trying to compare to the continuous case, it seemed interesting since $$\int \log \left(\sqrt{j}+1\right)\,dj=\sqrt{j}-\frac{j}{2}+(j-1) \log \left(\sqrt{j}+1\right)$$
Using Euler-MacLaurin summation formula
$$\log\left(\prod _{j=1}^n \left(\sqrt{j}+1\right)\right)=C+\frac{1}{2} n (\log (n)-1)+2 \sqrt{n}-\frac 14 \log(n)-$$ $$\frac{1}{6 \sqrt{n}}\left(1-\frac{1}{4 \sqrt{n}}+\frac{1}{20 n}-\frac{1}{140 n^2}+\frac{1}{120 n^2\sqrt n}+O\left(\frac{1}{n^3}\right)\right)$$
For this level of expansion $$C=\frac{\log (2)}{2}-\frac{88974215977}{87199580160}$$
For $n=100$, this truncated series gives $198.41718$ while the exact value is $198.41707$.