$\left(1+\frac{1}{3}\right)\left(1+\frac{1}{3^2}\right)\cdots\left(1+\frac{1}{3^n}\right)< \sqrt{e}$ is also true.

Use the fact that $\ln(1+x) < x - \dfrac {x^2}{2} + \dfrac {x^3}{3}$ for all $x > 0$. This can be proved considering the function $f(x) = x - \dfrac {x^2}{2} + \dfrac {x^3}{3} - \ln (1+x)$. We have $f'(x) = 1- x + x^2 - \dfrac {1}{1+x}$, $f''(x) = -1 + 2x + \dfrac {1}{(1+x)^2}$, $f'''(x) = 2 - \dfrac {2}{(1+x)^3} \geq 0$. So $f''$ is increasing, and $f''(0) = 0$. So $f'$ is increasing, and $f'(0) = 0$. So $f$ is increasing, and $f(0) = 0$. So $f(x) > 0$ for all $x> 0$. Then ${{\sum_{i=1}^n \ln (1 + \dfrac {1} {3^i}} ) < \sum_{i=1}^\infty \ln (1 + \dfrac {1} {3^i}} ) < \sum_{i=1}^\infty \dfrac {1}{3^i} - \dfrac {1}{2} \sum_{i=1}^\infty \dfrac {1}{9^i} + \dfrac {1}{3} \sum_{i=1}^\infty \dfrac {1}{27^i} = \dfrac {281}{624}$. Therefore ${\prod_{i=1}^n (1 + \dfrac {1} {3^i}} ) < \textrm{e}^{\dfrac {281}{624}} \approx 1.569 < \textrm{e}^{\dfrac {1}{2}} = \sqrt{\textrm{e}}$. See http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=608217&p=3616306#p3616306

arqady wrote: $\left(1+\frac{1}{3}\right)\left(1+\frac{1}{3^2}\right)\cdots\left(1+\frac{1}{3^n}\right)< \sqrt{e}$ is also true. By $ln(1+\alpha)<\alpha $ $(\alpha>0)$, $\left(1+\frac{1}{3}\right)\left(1+\frac{1}{3^2}\right)\cdots\left(1+\frac{1}{3^n}\right)=e^{ln\left(1+\frac{1}{3}\right)+ln\left(1+\frac{1}{3^2}\right)+\cdots+ln\left(1+\frac{1}{3^n}\right)}$ $<e^{\frac{1}{3}+\frac{1}{3^2}+\cdots+\frac{1}{3^n}}=e^{\frac{1}{2}(1-\frac{1}{3^n})}<e^{\frac{1}{2}}=\sqrt{e}.$