Infinite series that can be calculated! Find the value of the following sum:$$\sum_{n=1}^{\infty}\ln\left(1+x^{2^{n}}\right)$$
I wanted to use the fact that when one adds up logarithms the arguments multiply but then I got stuck. I don't know how to make it telescopic!
 A: I can see two ways of doing this.
($0< x < 1$ is assumed everywhere, otherwise the sun can't be calculated.) 
The first follows @achille hui's comment:
$$\begin{align*}\sum_{n=1}^{\infty}\ln\Big(1+x^{2^{n}}\Big) &= \sum_{n=1}^{\infty}\ln\Bigg(\frac{1-x^{2^{n+1}}}{1-x^{2^n}}\Bigg) \\
&= \ln\Bigg(\prod_{n=1}^{\infty}\frac{1-x^{2^{n+1}}}{1-x^{2^n}} \Bigg) \\
&= \ln \Bigg( \lim_{n \to \infty} \frac{1-x^{2^{n+1}}}{1-x^2} \Bigg) \\
&= \ln\bigg(\frac{1}{1 - x^2}\bigg)
\end{align*}$$
The second follows a now-deleted suggestion from the comments:
$$\begin{align*}\sum_{n=1}^{\infty}\ln\Big(1+x^{2^{n}}\Big) &= \ln \Bigg(\prod_{n=1}^{\infty} \Big(1 + x^{2^n}\Big)\Bigg) \\
&= \ln \Big(\Big(1 + x^2\Big)\Big(1 + x^4\Big)\Big(1 + x^8\Big)\dotsm \Big(1 + x^{2^n}\Big)\dotsm \Big) \\
&= \ln \big(1 + x^2 + x^4 + x^6 + x^8 + \dotsb + x^n + \dotsb\big) \\
&= \ln \bigg(\frac{1}{1 - x^2}\bigg)
\end{align*}$$
In case it's not clear, any even number can be uniquely made by adding some of the numbers from $\{2, 4, 8, 16, \dotsc\}$ and no odd number can be made, through a direct analogy to the binary representation of numbers.
I personally like the second method better because it's more elegant, intuitively appealing, and the parentheses get satisfactorily smaller towards the end.
