Is the infinite product of $-1 \times -1 \times -1 \times\dots = -i$? So I woke up this morning and I was thinking about the infinite product $-1 \times -1 \times -1 \times\dots$, and what it equals. I came to the conclusion that it equals $-i$. Alternatively stated,
$
{\displaystyle \prod_{i}^{\infty} (-1)} = -i
$
Here's how I reached this:
$${\prod_{i}^{\infty} (-1)} = e^{\ln({\displaystyle \prod_{i}^{\infty} (-1)})} = e^{\displaystyle \sum_{i}^{\infty}{\ln(-1)}}= e^{\displaystyle \sum_{i}^{\infty}{i\pi}}=e^{i\pi\displaystyle \sum_{i}^{\infty}{1}}$$
Now, here's where I'm a little hesitant. I want to say that, from $\zeta(0)=-\frac{1}{2}$, we can conclude that 
$$e^{i\pi\displaystyle \sum_{i}^{\infty}{1}} = e^{-\frac{1}{2}i\pi} = -i$$.
I have been told before that the sum $\displaystyle \sum_{i}^{\infty}{1}$ is not actually $-\frac{1}{2}$, but I'm not really sure why. It would seem that if this is the case, then my product would in fact not be $-i$. Though, I must say that $-i$ sort of makes sense, because multiplying complex numbers is essentially rotating them, and so rotating by $180$ every time will get you $180+180+180+...$ is the same as $180*(1+1+1+...)$ which is (if my premise is right) $180*(-\frac{1}{2})=-90$. $-90$ degrees on the complex plane turns out to be $-i$.
So my question is, is there a hole in my logic? I know what not accounting for $\zeta(0)=-\frac{1}{2}$, the sum $1+1+1+...$ is divergent, but taking that into account, can I say with confidence that $-1 \times -1 \times -1 \times\dots = -i$?

$$\prod_{n=1}^{\infty}(1+c)=\sum_{n=0}^{\infty}(2n)!/(n!)^2*(-c/4)^n=\frac{1}{\sqrt{(1+c)}}$$
The problem i've is that I found  $-1=e^{i\pi}$ why wouldn't it be $-1=e^{3i\pi}$?
And thuse according to your logic the answer can also be i.
