The awesome product

we split the product into 2 parts, let $exp(x)=e^x,H_n=\sum_{k=1}^n \dfrac{1}{k}$ .the product can be written as $\lim_{n\rightarrow \infty} \prod_{k=1}^n \dfrac{k}{k+11}\prod_{k=1}^n exp\left(\dfrac{11}{k}\right)$ The first product is just a basic telescoping series - product . the second one can be evaluated using the rules of exponents . $=\lim_{n\rightarrow \infty}\dfrac{11!}{(n+1)(n+2)...(n+11)}exp\left(\sum_{k=1}^n \dfrac{11}{n}\right)=\lim_{n\rightarrow \infty}\dfrac{11!}{(n+1)(n+2)...(n+11)}exp\left(11H_n\right)$ we know as n tends towards infinity, so does the harmonic series! But subtracting the log of n from it is finite! This motivates us to write this as $\lim_{n\rightarrow \infty}11!exp\left(11H_n-\ln((n+1)(n+2)...(n+11)))\right)\\=\lim_{n\rightarrow \infty}11!exp\left((H_n-\ln(n+1))+(H_n-\ln(n+2))+...(H_n-\ln(n+11)))\right)$ Remember $\lim_{n\rightarrow\infty}\left(H_n-\ln(n)\right)=\gamma\approx .5772$ we can also say $\lim_{n\rightarrow\infty} (H_n-\ln(n+1))=\lim_{n\rightarrow\infty} (H_{n+1}-\ln(n+1))=\gamma$ this is beacause $\lim_{n\rightarrow\infty} \dfrac{1}{n}=0$ so, it becomes $11!exp\left(11\gamma\right)\approx \boxed{22837737537}$