I think $(-e^{i\pi})(-e^{i\pi})(-e^{i\pi})(-e^{i\pi})(-e^{i\pi}).. \ne (-e^{i\pi})(-e^{i\pi})(-e^{i\pi})(1+1)$.

The infinite product $(−e^{iπ})(−e^{iπ})(−e^{iπ})...$ does not exist.

I don't see why one could simply forget $(1+1)$ when "doing it infinitley"

When you write "There is no scope ...", what in the world do you mean?

You should have written "We can do it infinitely. $$\implies(-e^{i\pi})(-e^{i\pi})\cdots(-e^{i\pi})(-e^{i\pi})(1+1)$$"

How can $1+1$ implie $-e^{i\pi}-e^{i\pi}$? That logically doesnt make sense

@Ahan please read this: en.m.wikipedia.org/wiki/Statement_(logic) $1+1$ simply isnt even a statement

"Nothing more than infinite" is not quite right. Infinity has strange properties like, $\infty + 1 = \infty$ and when $c > 0, c\cdot \infty = \infty$ That still doesn't mean you can throw away factors which occur after $1^\infty $

The other commenters are trying to help you. First as tina mentions, you should be putting the equals to symbol between your equations, not implications. Second, you are making the erraneous step of assuming that you can't write the $(1+1)$ at the end because you have already infinitely many things before it. But consider the following : just bring that to the front just after D (since multiplication here is commutative). Then you have $(1+1)\times(\text{ a bunch of things })$.

Just to drive home the point, we have the sum $1+1/2+1/4+1/8+\cdots$ and we can in fact do $(1+1/2+1/4+1/8+\cdots)+1$. This is completely valid. We just do it like $1$, then $1+1$, then $1+1/2+1$, then $1+1/2+1/4+1$ and so on.

@Tina That part is actually fine. $e^{i\pi}=-1 \implies -e^{i\pi}= 1$ so $-e^{i\pi} -e^{i\pi}= 1+1$

Hmmm, this may be the quickest I've seen a question closed.

For the record the problem is from step H to I.

Actually, it is.

@nickalh but that is not what he writes. How can anything implie $-e^{i\pi}-e^{i\pi}$, that isnt even a mathematical statement.