Proving $2+2^2/2+2^3/3+2^4/4+\cdots=0$ elementarily

In the first chapter of Gouvea's intro to $p$-adics, there's a heuristic argument that

$$ \frac{2}{1}+\frac{2^2}{2}+\frac{2^3}{3}+\frac{2^4}{4}+\cdots=0 \tag{$\ast$}$$

as $2$-adic numbers, since it's the Mercator series for $\ln(-1)$ and $2\ln(-1)=\ln(-1)^2=\ln1=0$.

(Like I said, heuristic.)

I assume $(\ast)$ can be proven by analyzing $\ln(1+x)$ as a function of $p$-adic numbers, but there's an exercise that says we can show $(\ast)$ by elementary means. But how? I feel like I've considered this question in the past before, but don't remember if I ever solved it.

We should be able to prove it's congruent to $0$ mod $2^N$ for any $N$. This automatically truncates the series to a finite sum, and all of the denominators divisible by $2$ are underneath numerators even more divisible by $2$, so it's well-defined. Perhaps we can split the sum into subsums of even and odd indices and establish a recursion?

Since $\lim_{n\to\infty}\nu_2\left(\frac{2^n}n\right)=\infty$, it suffices to prove that there are partial sums with arbitrarily high $\nu_2$. We adapt the first solution from here (pp. 9–10).

For $n$ even, we know from the Binomial Theorem that $$1=(-1)^n=(1-2)^n=\sum_{k=0}^n\binom nk(-2)^k.$$ This implies \begin{equation}\sum_{k=1}^n\frac 1n\binom nk(-2)^k=0\label{1}\tag{1}.\end{equation} Now, we can write $$\frac1n\binom nk=\frac{(n-1)\cdots(n-(k-1))}{k!}=\frac{(-1)^{k-1}(k-1)!+nm_{n,k}}{k!}=\frac{(-1)^{k-1}}k+n\frac{m_{n,k}}{k!},$$ for some integer $m_{n,k}$. We can therefore rewrite $(\ref{1})$ as \begin{equation}\sum_{k=1}^n\frac{2^k}k=n\sum_{k=1}^n\frac{m_{n,k}(-1)^k2^k}{k!}.\label{2}\tag{2}\end{equation} It is well-known that $\nu_2(k!)<k$: this is easy to prove using Legendre's Formula. Therefore, the $\nu_2$ of the RHS of $(\ref{2})$ is at least the $\nu_2$ of $n$. In particular, $$\nu_2\left(\sum_{k=1}^{2^n}\frac{2^k}k\right)\geq n,$$ and we're done. $\blacksquare$