Product with e

$\large \prod_{n=0}^\infty e^{\frac 1{n!}} = e^{\frac 1{0!}} e^{\frac 1{1!}} e^{\frac 1{2!}} \cdots = e^{\frac 1{0!}+\frac 1{1!} = \frac 1{2!} + \cdots} = e^e \approx \boxed{15.2}$

Represent e^x as a summation: $e^x=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$

Convert Summation to Product: $\sum_{n=0}^{\infty}\frac{x^{n}}{n!}=\ln\left(\prod_{n=0}^{\infty}e^{\frac{x^{n}}{n!}}\right)$

Replace Summation with e^x: $e^x=\ln\left(\prod_{n=0}^{\infty}e^{\frac{x^{n}}{n!}}\right)$

Solve for Product: $e^{e^{x}}=\prod_{n=0}^{\infty}e^{\frac{x^{n}}{n!}}$

Set x = 1: $e^{e}=\prod_{n=0}^{\infty}e^{\frac{1}{n!}}$

Round e^e: $e^e \approx 15.2$