Why is $e^{(x+2x+4x+...+2^{k-1}x)}=e^{((2^k-1)x)}$? Could anyone explain me why we have the following equality, $e^{(x+2x+4x+...+2^{k-1}x)}=e^{((2^k-1)x)}$? I don´t see what manipulation we did to get it
It's because $1 + 2 + 2^2 + 2^3 + \cdots + 2^{k-1} = 2^k - 1$.

And why is this true? Well, maybe an example will convince you why:
\begin{align*}
&amp;\quad\;\; 1 + (1 + 2 + 2^2 + 2^3 + 2^4 + 2^5) \\
&amp;= 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 \\
&amp;= 2^2 + 2^2 + 2^3 + 2^4 + 2^5 \\
&amp;= 2^3 + 2^3 + 2^4 + 2^5 \\
&amp;= 2^4 + 2^4 + 2^5 \\
&amp;= 2^5 + 2^5 \\
&amp;= 2^6
\end{align*}
So subtracting $1$ from both sides,
$$
1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 = 2^6 - 1.
$$
it is known that $1+a+a^2+...+a^m=(a^{m+1}-1)/(a-1)$. To see this cross multiply and telescope. So $x+2x+...+2^{k-1}x=x(1+2+2^2+...+2^{k-1})=x(2^k-1)$. The result follows.