$y=ce^x$ is the only solution to $y'=y$. This is known.
Let $k=\ln 2$
Let $S=\cdots+k^{-1}2^x+k^0 2^x+k^1 2^x+ k^2 2^x+\cdots$
Let $S'=\frac d{dx}S$
$S'=S$ so $S=ce^x$
$S=2^x[\cdots+k^{-1}+k^0+k^1+k^2+\cdots]$
Since $[\cdots+k^{-1}+k^0+k^1+k^2+\cdots]$ is not in terms of $x$ so we can set it equal to $a$, just some constant.
We can pull $a$ out of $S$ and divide it from both sides of $S=S'$ so we get $\frac d{dx}2^x=2^x$ which is not true.
What did I mess up on? I think it's when I pull the constant out and divide it on both sides, since it's infinitely large it might not work the same but I don't see why it wouldn't.
$$\sum _{k=-\infty}^{\infty } \log^k 2$$ does not converge. The positive indices give $$\sum _{k=0}^{\infty } \log^k 2=\frac{1}{1-\log 2}$$ but as $\dfrac{1}{\log 2}>1$ the sum $$\sum_{k=-\infty}^{-1}\log^k 2=\sum _{k=1}^{\infty } \left(\frac{1}{\log 2}\right)^k$$ doesn't converge so the "proof" is invalid