Evaluating $\prod_{n=1}^{\infty}\left(1+\frac{1}{2^n}\right)$

Is the product till infinity equal to $1$?

Certainly not! All the individual terms are greater than $1$. So if you multiply them together, you will always be increasing and cannot get back to $1$.

If no, what is the answer?

The product in question is $$ \prod_{n=1}^\infty (1 + x^n) $$ where $x = \frac12$. This product equals $$ \sum_{n=0}^\infty q(n) x^n $$ where $q(n)$ is the number of partitions of $n$ into distinct parts (each part $\ge 1$), and also equals $$ \prod_{n=1}^\infty \frac{1}{1 - x^{2n-1}} = \frac{\Phi(x^2)}{\Phi(x)} $$ (see Wikipedia), where here $\Phi$ is the Euler function, not to be confused with Euler's totient function. So your product is equal to $$ \boxed{\frac{\Phi(1/4)}{\Phi(1/2)} = 2.38423\ldots}. $$ I don't expect this can be simplified.

In the link you provided, the product seems to be $$\prod_{n=0}^{\infty} \left (1+\frac{1}{2^{2^n}}\right)$$

Note that $$(1-\frac{1}{2})\prod_{n=0}^{\infty} (1+\frac{1}{2^{2^n}})=\lim_{n \to \infty} 1-\frac{1}{2^{2^{n+1}}}=1$$ From the fact that $$(1+\frac{1}{2^{2^k}})(1-\frac{1}{2^{2^k}})=1-\frac{1}{2^{2^{k+1}}}$$

$$\prod_{n\geq 1}\left(1+\frac{1}{2^n}\right)=\exp\sum_{n\geq 1}\sum_{m\geq 1}\frac{(-1)^{m+1}}{m 2^{mn}}=\exp\sum_{m\geq 1}\frac{(-1)^{m+1}}{m(2^m-1)} $$ where the last series is (rapidly) convergent by Leibniz' test. It follows that the original product is between $\exp\left(\frac{121}{140}\right)$ and $\exp\left(\frac{3779}{4340}\right)$. By truncating the shown series at $m=30$ we get $$\prod_{n\geq 1}\left(1+\frac{1}{2^n}\right)\approx 2.384231029.$$