# infinite product of n=0 to infinity of (1+1/e^n)^(1/n!) $A=\prod_{n=0}^{\infty} \sqrt[n!]{1+\frac{1}{e^{n}}}$ $\log A=\sum_{n=0}^{\infty} \frac{1}{n!} \sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m}\left(\frac{1}{e^{n}}\right)^{m}$ $\log A=\sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m} \sum_{n=0}^{\infty} \frac{(e^{-m})^{n}}{n!}$ $\exp\left[\sum_{m=1}^{\infty} \frac{(-1)^{m+1}e^{e^{-m}}}{m}\right]=\prod_{n=0}^{\infty} \sqrt[n!]{1+\frac{1}{e^{n}}}$

# infinite product of n=1 to infinity of n * sin(1/n) $\prod_{n=1}^{\infty} n \sin(1/n)$

Since: $A=\prod_{n=1}^{\infty} n \sin(1/n)=\prod_{n=1}^{\infty} \prod_{m=1}^{\infty} \left(1-\frac{1}{n^{2}m^{2}\pi^{2}}\right)$ $\log(A) = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \log\left(1-\frac{1}{n^{2}m^{2}\pi^{2}}\right)$ $\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \sum_{v=1}^{\infty} \frac{(-1)^{v+1}}{v} \left[-\frac{1}{n^{2}m^{2}\pi^{2}}\right]^{v}$ $\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \sum_{v=1}^{\infty} \frac{(-1)^{2v+1}}{v} \left[\frac{1}{n^{2}m^{2}\pi^{2}}\right]^{v}$ $\sum_{v=1}^{\infty} \frac{(-1)^{2v+1}}{v}\frac{1}{\pi^{2v}} \left(\sum_{n=1}^{\infty} \frac{1}{n^{2v}}\right) \left(\sum_{m=1}^{\infty} \frac{1}{m^{2v}}\right)$ $\sum_{v=1}^{\infty} \frac{(-1)^{2v+1}}{v}\frac{\zeta(2v)^{2}}{\pi^{2v}}$

(which can easily be evaluated in wolfram alpha, and converges much faster than the original double product)

Therefore: $\prod_{n=1}^{\infty} n \sin(1/n)\approx 0.755363388518573214063...$