Let $f(n) = \Big(1+\frac{1}{3}\Big)\Big(1+\frac{1}{3^2}\Big)\cdots\Big(1+\frac{1}{3^n}\Big)$. Instead showing $f(n) < 2$, let's prove $f(n) < 2 - \frac{1}{3^n}$ using mathematical induction. When $n=1$, $f(1) = 1+\frac{1}{3} < 2 - \frac{1}{3}$. The claim holds. Suppose the claim holds when $n=k$, i.e. $f(k) < 2-\frac{1}{3^k}$. Then when $n=k+1$, \begin{align} f(k+1) &= f(k)\Big(1+\frac{1}{3^{k+1}}\Big)\\ &< \Big(2-\frac{1}{3^k}\Big)\Big(1+\frac{1}{3^{k+1}}\Big)\\ &= 2 - \frac{1}{3^k}+\frac{2}{2^{k+1}}-\frac{1}{3^{2k+1}}\\ &< 2 - \frac{1}{3^k} + \frac{2}{3^{k+1}}\\ &= 2 - \frac{1}{3^{k+1}} \end{align} Therefore, by the principle of mathematical principle, it always hold that $$f(n) < 2 - \frac{1}{3^n} < 2$$