From $1$ to infinit- $e$

$\displaystyle \begin{aligned} P & = \prod_{n \ = \ 1}^{\infty} \sqrt[4n^2 - 2n]{e} \\ & = \prod_{n \ = \ 1}^{\infty} e^{\frac{1}{4n^2 - 2n}} \\ & = \exp\left(\sum_{n \ = \ 1}^{\infty} \frac{1}{4n^2 - 2n}\right) \\ & = \exp\Bigg[\sum_{n \ = \ 1}^{\infty} \left( \frac{1}{2n - 1} - \frac{1}{2n} \right)\Bigg] \\ & = \exp\left(\sum_{n \ = \ 1}^{\infty} \frac{(- 1)^{n + 1}}{n}\right) \\ & = \exp\left(\ln 2\right) \\ & = \boxed{2} \end{aligned}$

This problem was the product of my curiousity:

I pondered $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+-...=\ln(2)$ :

$e^{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+-...}=2$

$e^1 \times e^{-\frac{1}{2}} \times e^{\frac{1}{3}} \times e^{-\frac{1}{4}}... =2$

$\exp(1) \times \frac{1}{\exp({\frac{1}{2})}} \times \exp({\frac{1}{3}}) \times \frac{1}{\exp({\frac{1}{4}})}... =2$

$\frac{\exp(1)}{\exp({\frac{1}{2}})} \times \frac{\exp(\frac{1}{3})}{\exp({\frac{1}{4}})} \times \frac{\exp(\frac{1}{5})}{\exp({\frac{1}{6}})} \times ... = 2$

$\exp({1-\frac{1}{2}}) \times \exp({\frac{1}{3}-\frac{1}{4}}) \times \exp({\frac{1}{5}-\frac{1}{6}}) \times ... = 2$

$\displaystyle \prod_{n=1}^\infty \exp(\frac{1}{2n-1}-\frac{1}{2n})=2$

Now, $\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{4n^2-2n}$ so

$\displaystyle \prod_{n=1}^\infty \exp(\frac{1}{4n^2-2n})=2$

and finally $\large \displaystyle \prod_{n=1}^\infty \sqrt[(4n^2-2n)]{e}= \boxed{2}$