<s>[INST] Proof that $ \frac{2^{-x}-1}{x} = \sum_{n=0}^{\infty} \frac{ (-1)^{n+1}x^n(\ln2)^{n+1}}{(n+1)!} $ 
If
$$2^x = \sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!} \hspace{1cm} \forall x \in \mathbb{R},$$
Proof that:
$$ \frac{2^{-x}-1}{x} =  \sum_{n=0}^{\infty} \frac{ (-1)^{n+1}x^n(\ln2)^{n+1}}{(n+1)!}  $$

I did the following:
\begin{align*}
&2^x = \sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!} \\
\Rightarrow \quad & 2^{-x} = \frac{1}{\sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!}} \\
\Rightarrow \quad & 2^{-x}-1 = \frac{1}{\sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!}} -1 \\
\Rightarrow \quad & \dfrac{2^{-x}-1}{x}  = \frac{\frac{1}{\sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!}} -1}{x} \\
\Rightarrow \quad & \frac{2^{-x}-1}{x}  = \frac{1-\sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!}}{x\sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!}}
\end{align*}
From this part I don't know how to continue
 [/INST] We have
$$2^x = \sum_{n=0}^{\infty} \dfrac{(x\ln(2))^n}{n!}$$
$$2^{-x} = \sum_{n=0}^{\infty} \dfrac{(-x)^n(\ln(2))^n}{n!}$$
$$2^{-x}-1 = -1+\sum_{n=0}^{\infty} \dfrac{(-1)^nx^n(\ln(2))^n}{n!}=\sum_{n=0}^{\infty} \dfrac{(-1)^{n+1}x^{n+1}(\ln(2))^{n+1}}{(n+1)!}$$
$$\frac{2^{-x}-1}x =\sum_{n=0}^{\infty} \dfrac{(-1)^{n+1}x^{n}(\ln(2))^{n+1}}{(n+1)!}$$
</s>