$y=ce^x$ 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.