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
 A: We have
$$2^x = \sum_{n=0}^{\infty} \dfrac{(x\ln(2))^n}{n!}$$
$$2^{-x} = \sum_{n=0}^{\infty} \dfrac{(-x)^n(\ln(2))^n}{n!}$$
$$2^{-x}-1 = -1+\sum_{n=0}^{\infty} \dfrac{(-1)^nx^n(\ln(2))^n}{n!}=\sum_{n=0}^{\infty} \dfrac{(-1)^{n+1}x^{n+1}(\ln(2))^{n+1}}{(n+1)!}$$
$$\frac{2^{-x}-1}x =\sum_{n=0}^{\infty} \dfrac{(-1)^{n+1}x^{n}(\ln(2))^{n+1}}{(n+1)!}$$
A: $$2^x = \sum_{n=0}^{\infty} \dfrac{(x\ln(2))^n}{n!} $$
Proof:
$$2^{-x} = \sum_{n=0}^{\infty} \dfrac{(-x\ln(2))^n}{n!} $$
$$\Rightarrow 2^{-1x}  = 1 +\sum_{n=1}^{\infty} \dfrac{(-1)^n(x\ln(2))^n}{n!} $$
$$\Rightarrow 2^{-x} -1  = \sum_{n=1}^{\infty} \dfrac{(-1)^n(x\ln(2))^n}{n!} $$
$$\Rightarrow 2^{-x} -1  = \sum_{n=0}^{\infty} \dfrac{(-1)^{n+1}(x\ln(2))^{n+1}}{(n+1)!} $$
$$\Rightarrow  \dfrac{2^{-x} -1}{x} = \sum_{n=0}^{\infty} \dfrac{(-1)^{n+1}x^n\ln(2)^{n+1}}{(n+1)!} $$
A: Your series can be written:
$$\sum_{n=0}^{\infty} \frac{\left(-1\right)^{n + 1} x^{n} \ln{\left(2 \right)}^{n + 1}}{\left(n + 1\right)!}=\sum_{n=0}^{\infty} \frac{x^{n} \left(- \ln{\left(2 \right)}\right)^{n + 1}}{\left(n + 1\right)!}$$
being $(-1)^{n+1}=-1$ forall $n\in\Bbb N$.
And rewriting
$$\sum_{n=0}^{\infty} \frac{x^{n} \left(- \ln{\left(2 \right)}\right)^{n + 1}}{\left(n + 1\right)!}=\sum_{n=0}^{\infty} \frac{\left(- x \ln{\left(2 \right)}\right)^{n + 1}}{x \left(n + 1\right)!}$$
Shift the series by $1$ you obtain:
$$\sum_{n=0}^{\infty} \frac{\left(- x \ln{\left(2 \right)}\right)^{n + 1}}{x \left(n + 1\right)!}=\sum_{n=1}^{\infty} \frac{\left(- x \ln{\left(2 \right)}\right)^{n}}{x n!}=\sum_{n=1}^{\infty} \frac{\left(- x \ln{\left(2 \right)}\right)^{n}}{x n!}= \frac{\displaystyle\sum_{n=1}^{\infty} \frac{\left(- x \ln{\left(2 \right)}\right)^{n}}{n!}}{x}$$
(remember that $x$ it is a constant).
After,
$$\frac{\sum_{n=1}^{\infty} \frac{\left(- x \ln{\left(2 \right)}\right)^{n}}{n!}}x=\frac{\left(\sum_{n=0}^{\infty} \frac{\left(- x \ln{\left(2 \right)}\right)^{n}}{n!} + \sum_{n=0}^{0} - \frac{\left(- x \ln{\left(2\right)}\right)^{n}}{n!}\right)}{x}=\frac{\sum_{n=0}^{\infty} \frac{\left(- x \ln{\left(2 \right)}\right)^{n}}{n!}-1}{x}=\frac{e^{- x \ln{\left(2 \right)}}-1}{x}$$
(you have an exponential series)
i.e.
$$\frac{e^{- x \ln{\left(2 \right)}}-1}{x}=\frac{2^{-x}-1}{x}$$
